Chapter 25: Capacitance - Farmingdale State College · 25.2 The Parallel Plate Capacitor A parallel...

“Most of the fundamental ideas of science areessentially simple, and may, as a rule, be expressed ina language comprehensible to everyone.”

Albert Einstein

25.1 IntroductionWhenever two nearby conductors of any size or shape, carry equal and oppositecharges, the combination of these conducting bodies is called a capacitor. Becausethe isolated conducting bodies have equal but opposite charges on them, an electricfield exists in the space between them. The importance of the capacitor lies in thefact that energy can be stored in the electric field between the two conducting bodies.Some simple examples of capacitors are shown in figure 25.1.

- - - - - - - - - - - - - -

E

++ + + + + +++

E

(a) +q

+

+

+

+

++

+ − −−

− −

q−

E

E

EE

(b)

Ε

q−q+

−−

−−−

Ε

Ε

Ε

−

−

+

+

+ ++ +

++

(c)

Figure 25.1 Some simple capacitors.

Figure 25.1(a) is a parallel plate capacitor which consists of two metal platesseparated by a distance d. A positive charge, +q, is placed on one of the plates, let ussay the top one, and a charge −q is placed on the bottom conducting plate.Neglecting any edge corrections, there is a uniform electric field E between the twocharged plates.

Figure 25.1(b) is a coaxial cylindrical capacitor. As the name implies, itconsists of two coaxial cylinders. The inner cylinder has a negative charge −q placedon it, while a positive charge +q is placed on the outer cylinder. The electric fieldfills the space between the cylinders as shown in figure 25.1(b).

A concentric spherical capacitor is shown in figure 25.1(c). The inner spherehas a negative charge −q placed on it, and a positive charge +q is placed on theouter sphere. A spherical electric field exists between the two conducting spheres. Inall of these cases, energy is stored in the electric field between the plates.

In the next few sections we will go into more detail on these capacitors. Wewill see that the study of capacitors is an excellent application of the concepts of thepotential we established in chapter 21, and Gauss’s law in chapter 22.

Chapter 25: Capacitance

25-1

25.2 The Parallel Plate CapacitorA parallel plate capacitor is connected to a battery as shown in figure 25.2(a).Although the actual charge carriers in a metal are electrons, we will continue withthe convention introduced in your college physics course that it is the positivecharges that are moving in the circuit. Recall that a flow of negative charges in onedirection is equivalent to a flow of positive charges in the opposite direction. Hence,

++ + + + + +++ I

II

III

E

E

Ad

Ad

E= 0

E= 0− − − − − − − − −

(c)

V

S

d

A

q−

q+

(a)- - - - - - - - - - - - - -

++ + + + + +++

E

V0

V1

V3V2

V4

V5

dl

d

(b)

Figure 25.2 The parallel plate capacitor.

using the concepts of conventional current, positive electric charge flows from thebattery of potential V to the left hand plate of the capacitor when the switch S isclosed. The charge distributes itself over the left hand plate. This positive charge onthe left hand plate induces an equal but negative charge on the right hand plate.The positive charge on the originally neutral right plate is pushed into the battery.The net result of applying the battery to the capacitor is that a charge of +q isdeposited on the left plate and −q on the right plate and a uniform electric field hasbeen set up between the plates. In effect, the battery has supplied energy to movepositive charges from the right plate and transferred them, through the battery, tothe left plate.

The relationship between the potential V across the plates, the electric field Ebetween the plates, and the plate separation d can be seen in figure 25.2(b). To findthe potential difference between the parallel plates we use equation 23.21.

(23.21)VB − VA = −¶ E * dl

Remember that equation 23.21 was derived as the work done per unit charge as acharge was moved up a potential hill from a point A at a lower potential to a pointB, at a higher potential. In figure 25.2(b) the work is done as we move along a pathfrom the lower potential at the negative plate to the higher potential at the upperplate. Hence the path dl points upward while the direction of the electric field vectorE points downward into the lower plate. Therefore the angle θ between E and dl is1800, and equation 23.21 becomes

Chapter 25 Capacitance

25-2

(25.1)VB − VA = −¶ E * dl = −¶ Edl cos 1800 = −¶ Edl(−1) = ¶ Edl

We defined the zero of potential to be at the negative plate, that is, VA = 0, and thepoint B is the positive plate and hence we let VB = V the potential between the twoplates. Also, as shown in chapter 19, the electric field E between the parallel platesis a constant and can come out of the integral sign. The difference in potentialbetween the plates now becomes

(25.2)V = ¶0d

Edl = E(d − 0)

and the potential between the plates becomes

(25.3)V = Ed

Let us determine the electric field E between the two oppositely chargedconducting plates, shown in figure 25.2(c), by Gauss’s law. Note that the solution forE is the same solution we found in section 4.10. We start by drawing a Gaussiancylinder, as shown in the diagram. One end of the Gaussian surface lies within oneof the parallel conducting plates while the other end lies in the region between thetwo conducting plates where we wish to find the electric field. Gauss’s law is givenby equation 4-14 as

(22.14)E = “ E * dA =qo

The sum in equation 22.14 is over the entire Gaussian surface. We break the entiresurface of the cylinder into three surfaces. Surface I is the end cap on the top-side ofthe cylinder, surface II is the main cylindrical surface, and surface III is the end capon the bottom of the cylinder as shown in figure 25.2(c). The total flux Φ throughthe Gaussian surface is the sum of the flux through each individual surface. That is,

Φ = ΦI + ΦII + ΦIII (25.4)where

ΦI is the electric flux through surface IΦII is the electric flux through surface II

ΦIII is the electric flux through surface III

Gauss’s law becomes (25.5)E = “ E * dA = ¶

IE * dA + ¶

IIE * dA + ¶

IIIE * dA =

qo

Because the plate is a conducting body, all charge must reside on its outer surface,hence E = 0, inside the conducting body. Since Gaussian surface I lies within theconducting body the electric field on surface I is zero. Hence the flux throughsurface I is,


25-3

(25.6)I = ¶I

E * dA = ¶I

E dA cos = ¶I(0) dA cos = 0

Along cylindrical surface II, dA is everywhere perpendicular to the surface. E lies inthe cylindrical surface, pointing downward, and is everywhere perpendicular to thesurface vector dA on surface II, and therefore θ = 900. Hence the electric fluxthrough surface II is

(25.7)II = ¶II

E * dA = ¶II

E dA cos = ¶II

E dA cos 90o = 0

Surface III is the end cap on the bottom of the cylinder and as can be seen fromfigure 25.2(c), E points downward and since the area vector dA is perpendicular tothe surface pointing outward it also points downward. Hence E and dA are parallelto each other and the angle θ between E and dA is zero. Hence, the flux throughsurface III is

(25.8)III = ¶III

E * dA = ¶III

E dA cos = ¶III

E dA cos 0o = ¶III

E dA

The total flux through the Gaussian surface is equal to the sum of the fluxesthrough the individual surfaces. Hence, using equations 25.6, 25.7, and 25.8,Gauss’s law becomes

Φ = ΦI + ΦII + ΦIII

(25.9)E = 0 + 0 + ¶III

EdA =qo

But E is constant in every term of the sum in integral III and can be factored out ofthe integral giving

E = E ¶ dA =qo

But !dA = A the area of the end cap, hence

E A =qo

A is the magnitude of the area of Gaussian surface III and q is the charge enclosedwithin Gaussian surface III. Solving for the electric field E between the conductingplates gives

(25.10)E =qoA

Equation 25.10, as it now stands, describes the electric field in terms of the chargeenclosed by Gaussian surface III and the area of Gaussian surface III. If the surfacecharge density on the plates is uniform, the surface charge density enclosed byGaussian surface III is the same as the surface charge density of the entire plate.Hence the ratio of q/A for the Gaussian surface is the same as the ratio q/A for the


25-4

entire plate. Thus q in equation 25.10 can also be interpreted as the total charge qon the plates, and A can be interpreted as the total area of the conducting plate.Therefore, equation 25.10 is rewritten as

(25.11)E =qoA

where E is the electric field between the conducting plates and is given in terms of thecharge q on the plates, the cross-sectional area A of the plates, and the permittivity εo

of the medium between the plates. Substituting equation 25.11 into equation 25.3 gives

(25.12)V =qdoA

Upon solving for q (25.13)q = oA

d V

Notice from equation 25.13 that the charge q on the plate is directly proportional tothe potential V between the plates, which is of course supplied by the battery. Thegreater the battery voltage V, the greater will be the charge q on the plate; thesmaller the voltage V, the smaller the charge q.

Let us now look carefully at the term in parenthesis in equation 25.13. Noticethat it is a constant and is a function of the geometry of the capacitor. As you recallfrom chapter 20, εo is called the permittivity of free space and has approximately thesame value for air as for a vacuum. This term is a function of the medium betweenthe plates, which in this case is air. A, in equation 25.13, is the cross sectional areaof a plate of the capacitor and d is the separation between the two plates. Becauseall these terms are constant for a particular capacitor, they are set equal to a newconstant C, called the capacitance of the capacitor. Therefore, the capacitance of aparallel plate capacitor is

(25.14)C = oAd

The capacitance is thus a function of the geometry of the capacitor itself. The largerthe area A of the plates, the greater will be the value of the capacitance C. Thegreater the plate separation d the smaller will be the capacitance C. A parallel platecapacitor of any value can be obtained by proper selection of area, plate separation,and the medium between the plates. So far, our discussion has been limited to acapacitor with air between the plates. In chapter ?7 an insulating material will beplaced between the plates.

The introduction of the concept of the capacitance allows us to write equation25.13 in the more general form

(25.15)q = CV


25-5

The charge q on a capacitor is directly proportional to the potential V between theplates, and the constant of proportionality is the capacitance C of the particularcapacitor. The capacitance can then be defined in general, from equation 25.15, to be

(25.16)C =qV

The SI unit for capacitance is defined from equation 25.16 to be a farad Fwhere

1 farad = 1 coulomb/voltThis is abbreviated as

1 F = 1 C/V

This unit is named in honor of Michael Faraday (1791-1867), an English physicist.If a charge of one coulomb is placed on the plates and the potential differencebetween the plates is one volt, the capacitance is then defined to be one farad. Acapacitance of one farad is extremely large, and the smaller units of microfarads,µF, or picofarads, pF, are usually used.

1 µF = 10−6 F 1 pF = 10−12 F

Example 25.1

Capacitance of a parallel plate capacitor. A parallel plate capacitor consists of twometal disks, 5.00 cm in radius. The disks are separated by air and are a distance of4.00 mm apart. A potential of 50.0 V is applied across the plates by a battery. Find(a) the capacitance C of the capacitor, and (b) the charge q on the plate.

Solution

a. The area of the plate is

A = πr2 = π(0.0500 m)2 = 7.85 × 10−3 m2

The capacitance is found from equation 25.14 to be

C = oAd =(8.85 % 10−12 C2/N m2 )(7.85 % 10−3 m2 )

(4.00 % 10−3 m) C = 17.4 % 10−12 C2

N mN m

JC = 17.4 % 10−12 C

J/CJ/CV

FC/V

C = 17.4 × 10−12 F C = 17.4 pF


25-6

Note how the conversion factors have been carried through in the example to showthat the capacitance does indeed come out to have the unit of farads.

b. The charge on the plate is determined from equation 25.15 as

q = CV q = (17.4 × 10−12 F)(50.0 V)

q = 8.70 × 10−10 C

To go to this Interactive Example click on this sentence.

25.3 The Cylindrical CapacitorA cylindrical capacitor, consisting of two coaxial cylinders of radii ra and rb, is shownin figure 25.3(a). The capacitor has a length l. A charge −q is placed on the inner

III

IIIdA

EE

E

dA dA

rL

(c)

+q

+

+

+

+

++

+ − −−

− −

q−

E

E

EE

(a)

+q

+

+

+

+

++

+ − −−

− −

q−

E

E

ra

rb

(b)

r

dl

Figure 25.3 A cylindrical capacitor.

cylinder and + q on the outer cylinder. Hence, an electric field E emanates from theouter cylinder, points inward, and terminates on the inner cylinder as shown. Thecapacitance C of the cylindrical capacitor is found from equation 25.16 as

(25.16)C =qV

In order to determine the capacitance from equation 25.16 it is first necessaryto determine the potential V between the two cylinders. The potential differencebetween the two cylinders is found from equation 23.21 as

(23.21)VB − VA = −¶ E * dl


25-7

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX25-01.XLS

Remember that equation 23.21 was derived as the work done per unit charge as acharge was moved up a potential hill from a point A at a lower potential to a pointB, at a higher potential. In figure 25.3(b) the work is done as we move along a pathfrom the lower potential at the negative inner cylinder to the higher potential at theouter positive cylinder. Hence the path dl points outward while the direction of theelectric field vector E points into the inner cylinder. Therefore the angle θ betweenE and dl is 1800, and equation 23.21 becomes

VB − VA = −¶ E * dl = −¶ Edl cos 1800 = −¶ Edl(−1) = ¶ Edl

We will assume that the negative plate is at the ground potential and hence we willtake VA = 0. The potential difference between the plates is then V and is given by

(25.17)V = ¶ Edl

But the element of path dl is equal to the element of radius dr, and hence

(25.18)V = ¶ra

rb Edr

In order to evaluate equation 25.18 it is necessary to know the electric field Ebetween the cylinders. To do this we use Gauss’s law, equation 22.14, modified totake into account that the charge enclosed by the Gaussian surface is negative, thatis, the charge on the inner cylinder is −q, . That is,

(25.19)E = “ E * dA =−qo

To determine the electric field between the two conducting cylinders, ra < r < rb, wedraw a cylindrical Gaussian surface of radius r and length L between the innercylinder and the outer cylinder as shown in figure 25.3(c). We assume that thecharge is uniformly distributed and has a surface charge density σ. The integral inGauss’s law is over the entire Gaussian surface. As we did in chapter 4, we breakthe entire cylindrical Gaussian surface into three surfaces. Surface I is the end capon the left-hand side of the cylinder, surface II is the main cylindrical surface, andsurface III is the end cap on the right-hand side of the cylinder as shown in figure25.3(c). The total flux Φ through the entire Gaussian surface is the sum of the fluxthrough each individual surface. That is,

Φ = ΦI + ΦII + ΦIII (25.20)where

ΦI is the electric flux through surface IΦII is the electric flux through surface II

ΦIII is the electric flux through surface III


25-8

Hence, Gauss’s law becomes

(25.21)E = “ E * dA = ¶I

E * dA + ¶II

E * dA + ¶III

E * dA =−qo

Along cylindrical surface I, dA is everywhere perpendicular to the surface andpoints toward the left as shown in figure 25.3(c). The electric field intensity vector Elies in the plane of the end cylinder cap and is everywhere perpendicular to thesurface vector dA of surface I and hence θ = 900. Therefore the electric flux throughsurface I is

(25.22)I = ¶I

E * dA = ¶I

E dA cos = ¶I

E dA cos 90o = 0

Surface II is the cylindrical surface itself. As can be seen in figure 25.3(c), E iseverywhere perpendicular to the cylindrical surface pointing inward, and the areavector dA is also perpendicular to the surface and points outward. Hence E and dAare antiparallel to each other and the angle θ between E and dA is 1800. Hence, theflux through surface II is

(25.23)II = ¶II

E * dA = ¶II

E dA cos = ¶II

E dA cos 180o = ¶II

E dA(−1) = − ¶II

E dA

Along cylindrical surface III, dA is everywhere perpendicular to the surface andpoints toward the right as shown in figure 25.3(c). The electric field intensity vectorE lies in the plane of the end cylinder cap and is everywhere perpendicular to thesurface vector dA of surface III and therefore θ = 900. Hence the electric fluxthrough surface III is

(25.24)III = ¶III

E * dA = ¶III

E dA cos = ¶III

E dA cos 90o = 0

Combining the flux through each portion of the cylindrical surfaces we get

E = ¶I

E * dA + ¶II

E * dA + ¶III

E * dA =−qo

(25.25)E = 0 − ¶II

E dA + 0 =−qo

From the symmetry of the problem, the magnitude of the electric field intensity E isa constant for a fixed distance r from the axis of the inner cylinder. Hence, E can betaken outside the integral sign to yield

(25.26)−¶II

E dA = −E ¶II

dA =−qo


25-9

The integral !dA represents the sum of all the elements of area dA, and that sum isjust equal to the total area of the cylindrical surface. As we showed in section 4.6,the total area of the cylindrical surface can be found by unfolding the cylindricalsurface. One length of the surface is L, the length of the cylinder, while the otherlength is the unfolded circumference 2πr of the end of the cylindrical surface. Thetotal area A is just the product of the length times the width of the rectangle formedby unfolding the cylindrical surface, that is, A = (L)(2πr). Hence, the integral of dAis

!dA = A = (L)(2πr)

Thus, Gauss’s law, equation 25.26, becomes

(25.27)E ¶II

dA = E(L)(2r) =qo

Notice that the minus signs were on both side of the equation in Gauss’s law,equation 25.26, and now have been canceled out. Solving for the magnitude of theelectric field intensity E we get

(25.28)E =q

2orL

Equation 25.28, as it now stands, describes the electric field in terms of the chargeenclosed by Gaussian surface II and the area of Gaussian surface II. If the surfacecharge density on the inner cylinder is uniform, the surface charge density enclosedby Gaussian surface II is the same as the surface charge density of the entire innercylinder. Hence the ratio of q/A for the Gaussian surface is the same as the ratio q/Afor the entire cylinder. Thus q in equation 25.28 can also be interpreted as the totalcharge q on the cylinder, and A can be interpreted as the total area of the innerconducting cylinder. Therefore, equation 25.28, the electric field between thecylinders, is rewritten as

(25.29)E =q

2orl

Placing equation 25.29 back into equation 25.18 we get

V = ¶ra

rb Edr = ¶ra

rb q20rl dr = ¶ra

rb q20l

drr

V =q

20l ln r|rarb =

q20l (ln rb − ln ra)

and the potential between the cylinders becomes

(25.30)V =q

20l ln rbra


25-10

The capacitance is now found, by placing equation 25.30 into equation 25.16, as

(25.31)C =qV =

qq

20l ln rbra

Simplifying, the capacitance of the coaxial cylinders becomes

(25.32)C = 2olln(rb/ra )

Example 25.2

A cylindrical capacitor. Find the capacitance of a cylindrical capacitor of radii ra =2.00 cm and rb = 2.50 cm. The length of the capacitor is 15.0 cm.

Solution

The capacitance C of the cylindrical capacitor is found from equation 25.32 as

C = 2olln(rb/ra )

C = 2(8.85 % 10−12 C2/N m2 )(0.15 m)ln[(0.025 m)/(0.020 m)]

C = 3.74 × 10−11 F = 37.4 pF


25.4 The Spherical CapacitorA spherical capacitor, consisting of two concentric spheres of radii ra and rb, is shownin figure 25.4(a). A charge −q is placed on the inner sphere and + q on the outersphere. Hence, an electric field E emanates from the outer sphere, points inward,and terminates on the inner sphere as shown. The capacitance C of the sphericalcapacitor is found from equation 25.16 as

(25.16)C =

qV

In order to determine the capacitance from equation 25.16 it is first necessaryto determine the potential V between the two spheres The potential differencebetween the two spheres is found from equation 23.21 as


25-11


Ε

q−q+

−−

−−−

Ε

Ε

Ε

−

−

+

+

+ ++ +

++

(a)

ra

rb

Εq−

q+ Ε

(c)

dA

r

q−

q+

Ε

Ε

Ε

(b)

dlΕ

Figure 25.4 A spherical capacitor.

(23.21)VB − VA = −¶ E * dl

Remember that equation 23.21 was derived as the work done per unit charge as acharge was moved up a potential hill from a point A at a lower potential to a pointB, at a higher potential. In figure 25.4(b) the work is done as we move along a pathfrom the lower potential at the negative inner sphere to the higher potential at theouter positive sphere. Hence the path dl points outward while the direction of theelectric field vector E points into the inner sphere. Therefore the angle θ between Eand dl is 1800, and equation 23.21 becomes

VB − VA = −¶ E * dl = −¶ Edl cos 1800 = −¶ Edl(−1) = ¶ Edl

We will assume that the inner negative sphere is at the ground potential and hencewe will take VA = 0. The potential difference between the two spheres is then V andis given by

(25.33)V = ¶ Edl

But the element of path dl is equal to the element of radius dr, and hence

(25.34)V = ¶ra

rb Edr

In order to evaluate equation 25.34 it is necessary to know the electric field Ebetween the spheres. To do this we use Gauss’s law, equation 22.14, modified totake into account that the charge enclosed by the Gaussian surface is negative, thatis, the charge on the inner sphere is −q. Thus,

(25.35)E = “ E * dA =−qo


25-12

To determine the electric field between the two conducting spheres, ra < r < rb, wedraw a spherical Gaussian surface of radius r between the inner sphere and theouter sphere as shown in figure 25.4(c). We assume that the charge is uniformlydistributed and has a surface charge density σ. The integral in Gauss’s law is overthe entire spherical Gaussian surface. As can be seen in figure 25.4(c), E iseverywhere perpendicular to the spherical surface pointing inward, and the areavector dA is also perpendicular to the spherical surface and points outward. HenceE and dA are antiparallel to each other and the angle θ between E and dA is 1800.Hence, the flux through the Gaussian surface is

(25.36) = ¶ E * dA = ¶ E dA cos = ¶ E dA cos 180o = ¶ E dA(−1) = −¶ E dA =−qo

From the symmetry of the problem, the magnitude of the electric field intensity E isa constant for a fixed distance r from the center of the inner sphere. Hence, E can betaken outside the integral sign to yield

(25.37)−¶ E dA = −E ¶ dA =−qo

The integral !dA represents the sum of all the elements of area dA, and that sum isjust equal to the total area of the spherical surface. But the area of a sphericalsurface is 4πr2. Hence, the integral of dA is

!dA = A = 4πr2 (25.38)

Thus, Gauss’s law, equation 25.37, becomes

(25.39)E ¶ dA = E(4r2) =qo

Notice that the minus signs were on both side of the equation in Gauss’s law,equation 25.37, and now have been canceled out. Solving for the magnitude of theelectric field intensity E we get

(25.40)E = 14o

qr2 =

kqr2

just as we would expect from our previous analyses in chapter 21. Placing equation25.40 back into equation 25.34 we get

V = ¶ra

rb Edr = ¶ra

rb kqr2 dr = ¶ra

rb kqr−2dr

V = kq(r−2+1)

(−2 + 1) |rarb = kq

(r−1)(−1) |ra

rb = −kq 1r |ra

rb = −kq( 1rb

− 1ra

) = kq( 1ra

− 1rb

)


25-13

and the potential between the two conducting spheres becomes

(25.41)V = kq( rb − rararb

)

The capacitance is now found from equations 25.16 and 25.41 as

(25.42)C =qV =

q

kq( rb − rararb )

= rarbk(rb − ra)

Simplifying, the capacitance of the concentric spheres becomes

(25.43)C = 4orarb(rb − ra)

Example 25.3

A spherical capacitor. Find the capacitance of a spherical capacitor of radii ra = 20.0cm and rb = 25.0 cm.

Solution

The capacitance C of the spherical capacitor is found from equation 25.43 as

C = 4orarb(rb − ra)

C =4(8.85 % 10−12 C2/N m2 )(0.20 m)(0.25 m)

(0.25 m) − (0.20 m)C = 1.11 × 10−10 F = 111 pF


Example 25.4

Capacitance of an isolated sphere. Find the capacitance of a charged isolated sphereof 35.0 cm radius.

Solution

The capacitance of an isolated sphere can be found by equation 25.43 by assumingthat the outer sphere is at infinity, i.e., rb = ∞. Hence,

C = 4orarb(rb − ra)


25-14


Dividing both numerator and denominator by rb we get

C = 4ora

( rbrb − ra

rb )= 4ora

(1 − rarb )

Letting rb approach infinity, this becomes

C = 4ora

(1 − ra∞ )

The capacitance of an isolated sphere becomes

(25.44)C = 4ora

C = 4π(8.85 × 10−12 C2 )(0.35 m) (N m2)C = 3.89 × 10−11 F


25.5 Energy Stored in a CapacitorAs shown in figure 25.2(a), using the concept of conventional current, the net effect ofcharging a capacitor by a battery is to take a positive charge q from the right plate,move it through the battery, and deposit it on the left plate. The work done by thebattery in moving the charge from the negative plate, or ground plate, to the positiveplate is converted to electrical potential energy of the charge. The mechanicalanalogue is that of a person who does work by lifting a bowling ball from the floor toa table, where the ball now has potential energy with respect to the floor. Since thecharge creates the electric field between the plates, this energy associated with thecharge may be viewed as residing in the electric field between the plates. Thus, theenergy stored in the capacitor is equal to the work done to charge the capacitor. Thework done by the battery in moving a charge q through the battery is

W = qV (25.45)

But in charging the capacitor the rate at which work is done is not a constant. Atthe instant that the switch in figure 25.2(a) is closed, the initial potential Vi acrossthe capacitor is zero. A small charge +qi is then placed on the left hand plate, whichthen induces the charge −qi on the right plate. An electric field E1 is establishedbetween the plates and a potential V1 appears across the plates. When the next


25-15


charge q1 is brought to the left plate, an amount of work W1 = q1V1 must be done bythe battery. With the new charge on the left plate, qi + q1, a new electric field E2 isestablished between the plates, and a new potential V2, which is greater than V1

appears across the plates. When the next charge q2 is brought to the left plate, thework done is W2 = q2V2. Since V2 is greater than V1, the work W2, must be greaterthan W1. With the new charge on the left plate, qi + q1 + q2, a new electric field E3 isestablished between the plates, and a new potential V3, which is greater than V2,appears across the plates. When the next charge q3 is brought to the left plate, thework done is W3 = q3V3. However, because V3 is greater than V2, the work done, W3 isgreater than the work W2. It is obvious that a different amount of work must bedone to move each charge to the plate, because as each charge is placed on the plate,a new potential appears across the plates and the product of qV will be different foreach charge. Hence, the total work necessary to charge the capacitor is equal to thesum, and hence integral, of all the products of qiVi for an extremely large number ofcharges. The problem can be solved by noting that the small amount of work dWthat is done in placing a small element of charge dq on the capacitor when there is apotential V across the plates is given by

dW = V dq (25.46)

The work done in charging the capacitor is the sum or integral of all these elementsof work and becomes

(25.47)W = ¶ dW = ¶ Vdq

but from equation 25.15, q = CV, and therefore V = q/C. Equation 25.47 thenbecomes

W = ¶ Vdq = ¶ qC dq

and upon integrating we get

(25.48)W =q2

2C

Equation 25.48, gives the energy that is stored in the electric field of the capacitor.Because the energy stored in the capacitor is equal to the work done to charge thecapacitor, the letter W will now be used to designate the energy stored in a capacitor,since the letter E, usually associated with energy, is now being used for the electricfield intensity. Using equation 25.15, q = CV, the energy stored in a capacitor canalso be written as

(25.49)W =q2

2C =(CV)2

2C = C2V2

2C (25.50)W = 1

2 CV2

Rearranging equation 25.15 into the form, C = q/V, and substituting it into equation25.50, the energy stored in the capacitor can also be written as


25-16

(25.51)W = 12 CV2 = 1

2 (q/V)V2

(25.52)W = 12 qV

Equations 25.48, 25.50, and 25.52 are different ways of expressing the energy thatis stored in the capacitor. This stored energy can be related to the electric fieldintensity between the plates of a parallel plate capacitor by using equations 25.50,25.14, and 25.3 as

(25.53)W = 12 CV2 = 1

2oAd

(Ed)2

(25.54)W = 12 oAdE2

The product of A, the cross sectional area of the plates, and d, the separationof the plates, is the volume between the plates that the electric field occupies, andas such is the volume of the electric field. This allows us to define a quantity calledthe energy density UE as the energy per unit volume, i.e.,

(25.55)UE =total energy

volume (25.56)UE = W

Ad = 12oAdE2

Ad (25.57)UE = 1

2 oE2

The energy density of the electric field, equation 25.57, was derived for the parallelplate capacitor, but it holds true in general for any electric field. Notice that theenergy density depends upon the electric field intensity. It is therefore certainlyappropriate to think of the energy as residing in the electric field.

Example 25.5

The energy stored in a capacitor. Find the energy stored in the capacitor of example25.1.

Solution

The energy stored in the capacitor is given by equation 25.50 as

W = 1/2 CV2 = 1/2 (17.4 × 10−12 F)(50.0 V)2

W = 2.18 % 10−8 C V2

VJ/CV

W = 2.18 × 10−8 J



25-17

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX25-05,6.XLS

Example 25.6

The energy density in the electric field. Find the energy density in the electric fieldbetween the plates of the above parallel plate capacitor.

Solution

Since the energy stored in the capacitor, W, is already known, the energy density isfound from equation 25.55 as

UE = Wvolume = W

Ad = 2.18 % 10−8 J(7.85 % 10−3 m2 )(4.00 % 10−3 m)

UE = 6.94 × 10−4 J/m3

As a check, let us find the electric field between the plates of the capacitor and thenuse equation 25.57 to find the energy density. The electric field between the platesis found from equation 25.1 as

E = Vd = 50.0 V

4.00 % 10−3 m = 1.25 % 104 V/m

and the energy density from equation 25.57 as

UE = 1/2 εoE2 UE = 1/2 (8.85 × 10−12 C2 /N m2)(1.25 × 104 V/m)2

UE = 6.94 × 10−4 J/m3

which is the same energy density determined in the previous calculation.


25.6 Capacitors in SeriesIf there are several capacitors in a circuit, another capacitor can be found that isequivalent to the entire combination present. As an example consider figure 25.5(a),which is a schematic diagram of a circuit containing three capacitors in series. (Notethat the symbol used for a capacitor in a circuit diagram is two parallel lines,symbolic of the parallel plate capacitor.) Again using the concept of conventionalcurrent, that is, a flow of positive charges, the battery causes charge +q to flow tothe left-hand plate of capacitor C1. This charge induces a charge −q in the right-handplate of C1. In this induction process, positive charges that were originally on theright-hand plate of C1 are forced to move away from the right-hand plate. The only


25-18


V

2C 3CC1

Figure 25.5 Capacitors in series.

place for them to go is the left-hand plate of C2, where they now distributethemselves. They in turn induce a charge −q on the right-hand plate of C2. Thiscauses a charge of +q to appear on the left-hand plate of C3, which in turn induces acharge −q on the right-hand plate of C3. The positive charges that were on theright-hand plate of C3 are now returned to the negative terminal of the battery. Thenet effect, therefore, is that when capacitors are connected in series, each capacitorhas the same charge q on its plates.

The potential drop across each capacitor, given by equation 23.6, is

V1 = q (25.58) C1

V2 = q (25.59) C2 V3 = q (25.60) C3

The total potential V across the three capacitors is equal to the sum of the potentialdrops across each capacitor, that is,

V = V1 + V2 + V3

Substituting equations 25.58, 25.59, and 25.60 into this equation gives

V = q + q + q C1 C2 C3 or

(25.61)V = q 1C1

+ 1C2

+ 1C3

Dividing both sides of equation 25.61 by q gives

(25.62)Vq = 1

C1+ 1

C2+ 1

C3

A rearrangement of equation 23.6 for a single capacitor gives


25-19

V = 1 (25.63) q C

Just as we introduced the concept of an equivalent resistor in resistive circuits, wenow introduce an equivalent capacitor for a circuit containing capacitors, as seen infigure 25.5(b). From equations 25.62 and 25.63, the equivalent capacitance forcapacitors in series is given by

(25.64)1C = 1

C1+ 1

C2+ 1

C3

That is, capacitors C1, C2, and C3 can be replaced in the series circuit by the oneequivalent capacitor C. Hence, the reciprocal of the equivalent capacitance ofcapacitors in series is equal to the sum of the reciprocals of each capacitor.

Example 25.7

Capacitors in series. Three capacitors of 3.00 µF, 6.00 µF, and 9.00 µF are connectedin series to a 50.0-V battery. Find (a) the equivalent capacitance, (b) the charge oneach capacitor, (c) the voltage drop across each capacitor, (d) the energy stored ineach capacitor, and (e) the total energy stored in the circuit.

Solution

a. The equivalent capacitance, found from equation 25.64, is

1C = 1

C1+ 1

C2+ 1

C3

= 13.00 % 10−6 F + 1

6.00 % 10−6 F + 19.00 % 10−6 F = 6.11 % 105 F−1

C = 1.64 × 10−6 F = 1.64 µF

b. Because the capacitors are in series the same charge is on each capacitor and isfound by

q = CV = (1.64 × 10−6 F)(50.0 V)= 8.18 × 10−5 C

c. The voltage drop across each capacitor, found from equations 25.58, 25.59, and25.60, is

V1 = q = 8.18 × 10−5 C = 27.3 V C1 3.00 × 10−6 F

V2 = q = 8.18 × 10−5 C = 13.6 V C2 6.00 × 10−6 F

V3 = q = 8.18 × 10−5 C = 9.09 V C3 9.00 × 10−6 F


25-20

Note that V1 + V2 + V3 = 50.0 V, which is equal to the applied voltage of 50.0 V.

d. The energy stored in each capacitor, found from equation 25.48, is

W1 = 1 q2 = 1 (8.18 × 10−5 C)2 = 1.12 × 10−3 J 2 C1 2 (3.00 × 10−6 F)

W2 = 1 q2 = 1 (8.18 × 10−5 C)2 = 0.558 × 10−3 J 2 C2 2 (6.00 × 10−6 F)

W3 = 1 q2 = 1 (8.18 × 10−5 C)2 = 0.372 × 10−3 J 2 C3 2 (9.00 × 10−6 F)

Note that W1 + W2 + W3 = 2.05 × 10−3 J.

e. We can find the total energy stored in the circuit by using the equivalent circuitof figure 25.5(b) as

W = 1 CV2 = 1 (1.64 × 10−6 F)(50.0 V)2

2 2= 2.05 × 10−3 J

Note that it checks with the result from part d.


25.7 Capacitors in ParallelConsider the three capacitors connected in parallel in figure 25.6(a). Using theconcept of conventional current, the battery causes a positive charge q to flow

V

2C

3C

C1

Figure 25.6 Capacitors in Parallel.


25-21


toward the capacitors. When it arrives at the junction A, it divides up into threedifferent charges, q1, q2, and q3, which are then deposited on the left-hand plates ofcapacitors C1, C2, and C3, respectively. Thus, when capacitors are connected inparallel there is a different amount of charge deposited on the plates of eachcapacitor. These positive charges induce their negative counterparts on theright-hand plate. The positive charges that were originally on the neutralright-hand plate are returned to the battery. Because charge is conserved

q = q1 + q2 + q3 (25.65)

But the relation between the charge and potential difference across each capacitor isgiven by equation 23.6 as

q1 = C1V1

q2 = C2V2

q3 = C3V3

Substituting these values of charge in equation 25.65, gives

q = C1V1 + C2V2 + C3V3 (25.66)

However, since the battery is in parallel with each capacitor the potential drop acrosseach capacitor is the same, that is,

V = V1 = V2 = V3

Equation 25.66 therefore becomesq = (C1 + C2 + C3)V (25.67)

An equivalent circuit is now introduced, as shown in figure 25.6(b). For thisequivalent circuit

q = CV

If this equivalent capacitance C is equal to the capacitance of equation 25.67, thenthe two circuits are in fact equivalent. Therefore,

(25.68)C = C1 + C2 + C3

is the equivalent capacitance of capacitors in parallel. Hence, the equivalentcapacitance of capacitors in parallel is equal to the sum of the individualcapacitances.

Example 25.8

Capacitors in parallel. Three capacitors of 3.00 µF, 6.00 µF, and 9.00 µF areconnected in parallel across a battery of 50.0 V, as shown in figure 25.6(a). Find


25-22

(a) the equivalent capacitance, (b) the total charge on the equivalent capacitor,(c) the voltage drop across each capacitor, (d) the charge on each capacitor, (e) theenergy stored in each capacitor, and (f) the total energy stored in the circuit.

Solution

a. The equivalent capacitance, found from equation 25.68, is

C = C1 + C2 + C3

= 3.00 µF + 6.00 µF + 9.00 µF= 18.00 µF

Again note the difference between capacitors connected in series and parallel. Whenthe same three capacitors were connected in series in example 25.7, the equivalentcapacitance was then 1.64 µF, very much different from the 18.00 µF when they arein parallel.

b. The total charge on the equivalent capacitor is found from

q = CV= (18.00 × 10−6 F)(50.0 V)

= 9.00 × 10−4 C

This is the total charge drawn from the battery.

c. The voltage drop across each capacitor is found by inspection. Since the capacitorsare all in parallel with the battery, the voltage drop across them is justequal to the battery voltage. Therefore,

V1 = V2 = V3 = V = 50.0 V

d. The charge on each capacitor is found from

q1 = C1V1 = C1V = (3.00 × 10−6 F)(50.0 V) = 1.50 × 10−4 Cq2 = C2V2 = C2V = (6.00 × 10−6 F)(50.0 V) = 3.00 × 10−4 Cq3 = C3V3 = C3V = (9.00 × 10−6 F)(50.0 V) = 4.50 × 10−4 C

Note that q1 + q2 + q3 = 9.00 × 10−4 C, which is equal to the total charge q drawnfrom the battery, which we found in part b.

e. The energy stored in each capacitor is

W1 = 1 C1V2 = 1 (3.00 × 10−6 F)(50.0 V)2

2 2


25-23

= 3.75 × 10−3 J W2 = 1 C2V2 = 1 (6.00 × 10−6 F)(50.0 V)2

2 2= 7.50 × 10−3 J

W3 = 1 C3V2 = 1 (9.00 × 10−6 F)(50.0 V)2

2 2= 11.25 × 10−3 J

Note that the total energy stored in the capacitors when they are in parallel is

W1 + W2 + W3 = 22.5 × 10−3 J

which is much greater than when the same capacitors were connected in series.

f. The total energy stored in the circuit can be found from the equivalent circuit as

W = 1 CV2 = 1 (18.00 × 10−6 F)(50.0 V)2

2 2= 22.5 × 10−3 J

Again, note that W1 + W2 + W3 = W.


Example 25.9

Redistributing the charge. A 6.00-µF capacitor is momentarily connected to a 50.0-Vbattery. The battery is then disconnected and a 9.00-µF capacitor is connected inparallel to the 6.00-µF capacitor (figure 25.7). Find (a) the original charge on thefirst capacitor, (b) the charge on each capacitor after they are connected in parallel,and (c) the voltage across the combination.

Figure 25.7 Connecting a charged capacitor to an uncharged one.


25-24


Solution

a. The initial charge q1i on the first capacitor is found from

q1i = C1V1i = (6.00 × 10−6 F)(50.0 V)= 3.00 × 10−4 C

b. When the second capacitor is connected in parallel to the first, the initial chargeon the first capacitor q1i redistributes itself between C1 and C2, such that there isnow a final charge q1f on C1 and a final charge q2f on C2. By the law of conservationof electric charge

q1i = q1f + q2f

Because the two capacitors are now in parallel, the voltage across them is equal andis expressed as Vf. This final voltage is therefore,

Vf = q1f = q2f C1 C2

the ratio of the charge distribution can be found as

q1f = C1 = 6.00 µF = 2 q2f C2 9.00 µF 3or

q1f = 2 q2f

3

That is, the final charge on C1 is two-thirds of the final charge on C2. To find theexact amount of charge, we substitute this relation back into the law of conservationof charge to obtain

q1i = q1f + q2f = 2 q2f + q2f = 5 q2f

3 3Therefore, the final charge on C2 is

q2f = 3 q1i = 3 (3.00 × 10−4 C) = 1.80 × 10−4 C 5 5

and the final charge on C1 is

q1f = 2 q2f = 2 (1.80 × 10−4 C) = 1.20 × 10−4 C 3 3


25-25

The initial charge of 3.00 × 10−4 C redistributes itself such that there is now 1.20 ×10−4 C on C1 and 1.80 × 10−4 C on C2.

c. The voltage across the capacitors is found from

V1f = q1f = 1.20 × 10−4 C = 20.0 V C1 6.00 × 10−6 F


25.8 Combinations of Capacitors in Series and ParallelIf capacitors are connected in combinations of series and parallel, as shown in figure25.8(a), the charge on each capacitor can be found by the techniques of sections 25.6and 25.7. Since capacitor C2 is parallel to capacitor C3, its equivalent capacitance,

Figure 25.8 Capacitors in series and parallel and the equivalent circuit.

found from equation 25.68, isC23 = C2 + C3

= 6.00 µF + 9.00 µF= 15.00 µF

Figure 25.8(a) is now equivalent to figure 25.8(b), here C23 = 15.00 µF and is inseries with C1. The equivalent capacitance of capacitors C1 and C23 in series, foundfrom equation 25.64, is


25-26


1 = 1 + 1 C123 C1 C23

1 = 1 + 1 = 0.400 C123 3.00 µF 15.00 µF µFand

C123 = 2.50 µF

Figure 25.8(b) is now equivalent to figure 25.8(c), with C123 = 2.50 µF. The totalcharge released from the battery becomes

q = C123V= (2.50 × 10

−6 F)(50.0 V)= 1.25 × 10−4 C

Since C1 is in series with the battery this total charge is deposited on the plates ofC1. Therefore,

q1 = q = 1.25 × 10−4 C

The voltage drop across C1 is found as V1 = q1 C1

= 1.25 × 10−4 C 3.00 × 10−6 F

= 41.67 VThe voltage drop across C23 is

V23 = V − V1 = 50.0 V − 41.67 V = 8.33 V

Since C2 and C3 are in parallel, the voltages across each capacitor are equal to eachother and to the voltage V23. That is,

V2 = V3 = V23

The charge on capacitor C2 is found from

q2 = C2V2

= (6.00 × 10−6 F)(8.33 V) = 5.00 × 10−5 C

And the charge on capacitor C3 isq3 = C3V3

= (9.00 × 10−6 F)(8.33 V)= 7.50 × 10−5 C


25-27

Note that the charge displaced from capacitor C1 has been distributed to capacitorsC2 and C3 and that

q1 = 1.25 × 10−4 C = q2 + q3

= 5.00 × 10−5 C + 7.50 × 10−5 C= 1.25 × 10−4 C

as it must by the law of conservation of charge.The charge on other simple combinations of capacitors can be found using the

same techniques.

25.9 Capacitors with Dielectrics Placed between thePlatesA dielectric is an insulator or nonconductor of electricity. Its effects are defined interms of the ratio of the capacitance of a capacitor with a dielectric between the platesto the capacitance of a capacitor without a dielectric between the plates. As anexample consider the parallel plate capacitor C0 with air between the plates, shownin figure 25.9(a). A battery is momentarily connected to the plates of the capacitor,

q−q+

(a)

Vo

oC

q−q+

(b)

Cd

Vd

Figure 25.9 Placing a dielectric between the plates of a capacitor.

thereby leaving a charge q on the plates and establishing a potential difference V0

between them. The battery is then removed. An electrometer, which is essentially avery sensitive voltmeter, is connected across the plates and reads the potentialdifference V0. The relationship between the charge, the potential, and thecapacitance is, of course,

C0 = q (25.70) V0

A very interesting phenomenon occurs when an insulating material, called adielectric, is placed between the plates of the capacitor, as shown in figure 25.9(b).The voltage, as recorded by the electrometer, drops to a lower value Vd. Since thecharge on the plates remains the same (there is no place for it to go because thebattery was disconnected), the only way the potential between the plates can


25-28

change is for the capacitance to change. The capacitance with the dielectric betweenthe plates can be written as

Cd = q (25.71) Vd

where Cd is the capacitance of the capacitor and Vd is the potential between theplates when there is a dielectric between them. Dividing equation 25.71 by equation25.70 gives

(25.72)CdCo

=q/Vd

q/Vo= Vo

Vd

and (25.73)Cd

Co= Vo

Vd

Because it is observed experimentally that Vd is less than V0, the ratio V0/Vd isgreater than 1 and hence Cd must be greater than C0. Therefore, placing a dielectricbetween the plates of a capacitor increases the capacitance of that capacitor. Thisshould not come as too great a surprise since we saw earlier that the capacitance isa function of the geometrical configuration of the capacitor. Placing an insulatorbetween its plates is a significant change in the capacitor configuration.

The effect of the dielectric between the plates is accounted for by theintroduction of a new constant, called the dielectric constant κ and it is defined asthe ratio of the capacitance with a dielectric between the plates to the capacitancewith a vacuum between the plates, that is,

(25.74) = CdCo

The dielectric constant depends on the material that the dielectric is made of, andsome representative values are shown in table 25.1. Note that κ for air is very closeto the value of κ for a vacuum, which allows us to treat a great many of electricproblems in air as if they were in vacuum.

The capacitance of any capacitor with a dielectric is easily found from thejuxtaposition of equation 25.74 to

(25.75)Cd = Co

For example, the capacitance of a parallel plate capacitor is found from equations6-14 and 25.75 to be

(25.76)Cd = o Ad

Quite often a new constant is used, called the permittivity of the medium and isdefined as

(25.77) = o


25-29

∞3

34040

200

3024

21

1.000001.00059

2.23880.373.123.5

5.402.74.54.8262.22.3

2.94

VacuumAir Carbon tetrachlorideWaterPlexiglass PaperMicaAmberPyrex glassBakeliteEthyl alcohol Wax paper BenzeneRubber

Dielectric Strength (×××× 106 V/m)

Dielectric Constant(κκκκ)

Material

Table 25.1Dielectric Constant and Dielectric Strength

Example 25.10

Placing a dielectric material between the plates of a capacitor. A parallel platecapacitor with air between the plates has a capacitance of 3.85 µF. If mica is placedbetween the plates of the capacitor what will the new capacitance be?

Solution

The value of the dielectric constant for mica is found in table 25.1 as κ = 5.40. Thecapacitance with the dielectric placed between the plates is found from equation25.75 as

Cd = κCo

= 5.40 × 3.85 × 10−6 F= 2.08 × 10−5 F


Example 25.11

Permittivity of a dielectric. Find the permittivity of the dielectric material mica.


25-30


Solution

The permittivity of mica is found from equation 25.77 as

ε = κεo

= (5.40)(8.85 × 10−12 C2/N m2)= 4.78 × 10−11 C2/N m2


25.10 Atomic Description of a DielectricAs was mentioned in chapter 20, all atoms are electrically neutral. We can see thereason for this from figure 25.10(a). The positive charge is located at the center

Figure 25.10 Placing an atom in a uniform electric field.

of the atom. The electron revolves around the nucleus. Sometimes it is to the rightof the nucleus, sometimes to the left, sometimes it is above the nucleus, sometimesbelow. Because of this symmetry, its mean position appears as if it were at thecenter of the nucleus. Therefore, it appears as though there is a positive charge anda negative charge at the center of the atom. These equal but opposite charges givethe effect of neutralizing each other, and hence the atom does not appear to haveany charges and appears neutral. The electric field of the positive charge is radiallyoutward, whereas the average electric field of the negative charge is radiallyinward. These equal but opposite average electric fields have the effect ofneutralizing each other, and hence the atom also does not appear to have anyelectric field associated with it. Because of this basic symmetry of the atom, allatoms are electrically neutral.

If the atom is now placed in a uniform external electric field E0, we observe achange in this symmetry, as shown in figure 25.10(b). The electron of the atom findsitself in an external field E0 and experiences the force


25-31


F = qE0 = −eE0

Because the electronic charge is negative, the force on the electron is opposite to thedirection of the external field E0. When the electron is to the right of the nucleus,the external electric field pulls it even further to the right, away from the nucleus.When the electron is to the left of the nucleus, the external field again pulls ittoward the right, this time toward the nucleus. The overall effect of this externalfield is to change the orbit of the electron from a circle to an ellipse, as shown infigure 25.10(b). The mean position of the negative electron no longer coincides withthe positive nucleus, but because of the new symmetry is displaced to the right ofthe nucleus, as shown in figure 23.8(c) The apparently neutral atom has beenconverted to an electric dipole by the uniform external electric field E0.

When two or more atoms combine they form a molecule. Many moleculeshave symmetric charge distributions and therefore also appear electrically neutral.Some molecules, however, do not have symmetric charge distributions and thereforeexhibit an electric dipole moment. Such molecules are called polar molecules. A slabof material made from polar molecules is shown in figure 25.11(a). In general,

Figure 25.11 A polar dielectric.

the electric dipoles are arranged randomly. If the dielectric is placed in an externalelectric field, electric forces act on the dipoles until the dipoles become aligned withthe field, as shown in figure 25.11(b).

The net effect is that a dielectric material made of dipole molecules has thosedipoles aligned in an external electric field. A dielectric material not composed ofpolar molecules has electric dipoles induced in them by the external electric field, asin figure 25.10. These dipoles are in turn aligned by the external electric field.Therefore, all insulating materials become polarized in an external electric field.The overall effect is to have positive charges on the right of the slab in figure25.11(b) and negative charges to the left. These charges are not free charges, butare bound to the atoms of the dielectric. They are merely the displaced charges andnot a flow of charges.


25-32

With this digression into the atomic nature of dielectrics, we can now explainthe effect of a dielectric placed between the plates of a capacitor. Figure 25.12(a)shows a charged capacitor with air between the plates. A uniform electric field E0 is

Figure 25.12 Electric field between the plates of the capacitor.

found between the plates. A slab of dielectric material is now placed between theplates, as shown in figure 25.12(b), and completely fills the space between theplates. The uniform electric field E0 polarizes the dielectric by aligning the electricdipoles, figure 25.12(b). The net effect is to displace positive bound charges to theright of the slab and negative bound charges to the left of the slab. These inducedbound charges have an electric field Ei associated with them. The induced electricfield Ei emanates from the positive bound charges on the right of the dielectric andterminates on the negative bound charges on the left of the dielectric, and, as wecan see from figure 25.12(c), Ei is opposite to the original electric field of thecapacitor. Hence, the total electric field within the dielectric is

(25.78)Ed ==== E0 −−−− Ei

Therefore, placing a dielectric between the plates of a capacitor reduces the electricfield between the plates of the capacitor.

The potential difference V0 across the plates of the capacitor, when there is nodielectric between the plates, is given by equation 23.7 as

V0 = E0d (25.79)

While the potential difference Vd across the capacitor when there is a dielectricbetween the plates is

Vd = Edd (25.80)

Dividing equation 25.79 by equation 25.80 gives

V0 = E0 (25.81) Vd Ed


25-33

But from equation 25.73 the ratio of the capacitance with a dielectric to thecapacitance without a dielectric was

Cd = Vo (25.73) Co Vd

Combining this result with equation 25.81 gives

Cd = Vo = E0 Co Vd Ed

Recall that the ratio of Cd/C0 was defined in equation 25.74 as the dielectricconstant, κ. Therefore,

(25.82)CdC0

= V0Vd

= E0Ed

=

The effect of the dielectric between the plates is summarized from equation 25.82 asfollows: 1. The capacitance with a dielectric increases to

Cd = κCo (25.75)

2. The potential difference between the plates decreases to

(25.83)Vd = V0

and 3. the electric field between the plates decreases to

(25.84)Ed = E0

The value of the induced electric field Ei within the dielectric can be foundfrom equation 25.84 and 25.78 as

Ed = Eo − Ei = Eo κ

Solving for Ei gives Ei = E0 − E0

= E0 1 − 1

(25.85)Ei = E0 − 1

The presence of a dielectric between the plates of a capacitor not onlyincreases the capacitance of the capacitor, but it also allows a much larger voltageto be applied across the plates. When the voltage across the plates of an aircapacitor exceeds 3.00 × 106 V/m, the air between the plates is no longer aninsulator, but conducts electric charge from the positive plate to the negative plateas a spark or arc discharge. The dielectric is said to break down. The capacitor is


25-34

now useless as a capacitor since it now conducts electricity through it and is said tobe short circuited. The value of the potential difference per unit plate separationwhen the dielectric breaks down is called the dielectric strength of the medium,and is given in the second column of table 25.1. If mica is placed between the plates,the breakdown voltage rises to 200 × 106 V/m, a value that is 66 times greater thanthe value for air.

Example 25.12

A capacitor with a dielectric. A parallel plate capacitor, having a plate area of 25.0cm2 and a plate separation of 2.00 mm, is charged by a 100 V battery. The battery isthen removed. Find: (a) the capacitance of the capacitor, (b) the charge on the platesof the capacitor. A slab of mica is then placed between the plates of the capacitor.Find, (c) the new value of the capacitance, (d) the potential difference across thecapacitor with the dielectric, (e) the initial electric field between the plates, (f) thefinal electric field between the plates, (g) the initial energy stored in the capacitor,and (h) the final energy stored in the capacitor.

Solution

a. The original capacitance of the capacitor, found from equation 25.14, is

C0 = 0 Ad = 8.85 % 10−12 C2

N m225.0 cm2

0.002 m1 m

102 cm2

= 1.11 % 10−11 C2

N mN m

J = 1.11 % 10−11 C

J/CJ/CV

FC/V

= 1.11 % 10−11 F = 11.1 % 10−12 FCo = 11.1 pF

b. The charge on the plates of the capacitor, found from equation 25.15, is

q = CoVo = (11.1 × 10−12 F)(100 V)(C/V) ( F )

q = 11.1 × 10−10 C

This charge remains the same throughout the rest of the problem because thebattery was disconnected.

c. The value of the capacitance when a mica dielectric is placed between the platesis found from equation 25.75 and table 25.1 as

Cd = κCo


25-35

Cd = (5.40)(11.1 pF)Cd = 59.9 pF

d. The new potential difference between the plates is found from equation 25.83 as

Vd = Vo = 100 V = 18.5 V κ 5.40

This could also have been found by equation 25.71 as

Vd = q = 11.1 × 10−10 C = 18.5 V Cd 59.9 × 10−12 F

e. The initial electric field between the plates is found from equation 25.79 as

Eo = Vo = 100 V = 5.00 × 104 V/m d 0.002 m

f. The final electric field between the plates, when the dielectric is present, is foundfrom equation 25.84 as

Ed = Eo = 5.00 × 104 V/m = 9.25 × 103 V/m κ 5.40

This could also have been found from equation 25.80 as

Ed = Vd = 18.5 V = 9.25 × 103 V/m d 0.002 m

g. The initial energy stored in the capacitor is found from equation 25.50 as

Wo = 1/2 CoVo2 = 1/2 (11.1 × 10−12 F)(100 V)2

Wo = 5.55 × 10−8 C V2 ( J ) V (CV)

Wo = 5.55 × 10−8 J

h. The final energy stored in the capacitor when there is a dielectric between theplates is

Wd = 1/2 CdVd2 (25.86)

The energy can be computed directly from equation 25.86, but it is instructive to dothe following. The capacitance Cd can be substituted from equation 25.75 and the


25-36

potential difference can be substituted from equation 25.83 into equation 25.86 toyield

Wd = 12 CdVd

2 = 12

(C0 ) V0

2

= 1

12 C0V0

2

(25.87)Wd = W0

This is the energy stored in the capacitance with a dielectric between the plates. Substituting the numbers into equation 25.87 gives

Wd = Wo = 5.55 × 10−8 J κ 5.40

Wd = 1.03 × 10−8 J

The energy stored in the capacitor with a dielectric is less than without the dielectric.This, at first, may seem to be a violation of the law of conservation of energy, sincethere is a decrease in energy. But this decrease in energy can be accounted for bythe work done in moving the dielectric slab between the plates of the capacitor.


Example 25.13

Placing a dielectric between the plates. A 6.00 pF air capacitor is connected to a 100V battery (figure 25.13). A piece of mica is now placed between the plates of thecapacitor. Find: (a) the original charge on the plates of the capacitor, (b) the newvalue of the capacitance with the dielectric, (c) the new charge on the plates.

Figure 25.13 Placing a dielectric between the plates while maintaining a constantpotential between the plates.

Solution

a. The original charge on the plates is found from


25-37


qo = CoVo = (6.00 × 10−12 F)(100 V)(C/V) ( F )

qo = 6.00 × 10−10 C

b. The new value of the capacitance is found from

Cd = κCo = (5.40)(6.00 × 10−12 F) Cd = 32.4 × 10−12 F

Cd = 32.4 pF

c. In this example, the battery is not disconnected from the battery. Therefore thepotential across the plates remains the same whether the capacitor has a dielectricor not. Because the capacitance of the capacitor has changed, the charge on theplates must now change and is found from

qd = CdVd qd = (32.4 × 10−12 F)(100 V)

qd = 32.4 × 10−10 C

By keeping the potential across the plates constant, more charge was drawn fromthe battery. A clear distinction must be made between this example and theprevious one. In example 25.12 the charge remained constant because the batterywas disconnected, and the potential varied by the introduction of the dielectric. Inthis example, the potential remained a constant, because the battery was notdisconnected, and the charge varied with the introduction of the dielectric.


B. Dielectrics and Coulomb’s Law. The electric field in a vacuum was defined as E = F/qo in equation 21.1. Calling theelectric field when no dielectric is present Eo, equation 21.1 becomes

Eo = F (25.88) q

But if a dielectric medium is present the electric field within the dielectric was givenby equation 25.84 as

Ed = Eo (25.84) κ

Replacing equation 25.88 into equation 25.84 gives


25-38


Ed = F (25.89) κq

Comparing equations 25.89 with equation 25.88 prompts us to define the electricfield in a dielectric as the ratio of the force acting on a particle within the dielectricto the charge, i.e.,

Ed = Fd (25.90) q

Comparing equations 25.89 and 25.90 allows us to define the force on a particle in adielectric as

(25.91)Fd = F

Using Coulomb’s law, we can now find the force between point charges in adielectric medium as

Fd = F = 1

40q1q2

r2

Recall from equation 25.77 that the permittivity of the medium is defined as

ε = κε0

In general then, Coulomb’s law should be expressed as

(25.92)F = 14

q1q2

r2

As the special case for a vacuum κ = 1, and equation 25.92 reduces to

F = 140

q1q2

r2

the form of Coulomb’s law used in chapter 20.As an example of the effect of the dielectric on the force between charges

consider the salt sodium chloride, NaCl, which is common table salt. Although NaClconsists of a lattice structure of Na+ and Cl− ions, let us just consider the electricforce between one Na+ ion, and one Cl− ion separated by a distance of 1.00 × 10−7 cm.The force between the ions is given by Coulomb’s law as

F0 = 140

q1q2

r2

= 9.00 % 109 N m2

C2(1.60 % 10−19 C)2

(1.00 % 10−9 m)2

= 2.30 × 10−10 N


25-39

If this salt is now placed in water the force between the ions becomes

Fd = 140

q1q2

r2

= 9.00 % 109

(80.0)N m2

C2(1.60 % 10−19 C)2

(1.00 % 10−9 m)2

= 0.0288 × 10−10 NThis could also be found as

Fd = F κ

= 2.30 × 10−10 N 80.0

= 0.0288 × 10−10 N

The force between the ions in the water solution has decreased by a factor of 1/80.This is why NaCl readily dissolves in water. The force between the ions becomessmall enough for simple thermal energy to pull the ions apart. In fact, mostchemicals are soluble in water because of this high value of 80 for the dielectricconstant. If the NaCl salt is placed in benzene, κ = 2.3, the force between the Na+

and Cl−−−− ions is not reduced enough to allow the ions to disassociate, and hence NaCldoes not dissolve in benzene.

The water molecule, H2O, is composed of two atoms of hydrogen and one atomof oxygen. The molecular configuration is not spherically symmetric, but is ratherasymmetric, as shown in figure 25.14(a). The oxygen atom has picked up two

Figure 25.14 The water molecule.

electrons, one from each hydrogen atom, and is therefore charged doubly negative.Each hydrogen atom, having lost an electron to the oxygen, is positive. The meanposition of the positive charge does not coincide with the location of the negativecharge but lies below it in the diagram. Hence, the water molecule behaves as anelectric dipole, as seen in figure 25.14(b), and it is a polar molecule. When amaterial, such as the sodium chloride, is placed in water, these water dipolessurround the Na+ and Cl− ions, as shown in figure 25.15. The negative ends of thewater molecule are attracted to the positive ions, whereas the positive end of thewater molecules are attracted to the negative ion. These dipoles that surround the


25-40

Figure 25.15 Water dipoles surround ions placed in water.

ions, shield the ions and have the effect of decreasing the effective charge of the ionsand hence decrease the coulomb force between them. Liquids with large values of κare the most effective in decreasing the force between the separated ions and aretherefore the best solvents for ionic crystals.

25.11 Charging and Discharging a CapacitorCharging a CapacitorUp to now in our discussion of capacitors, we saw their effects in different types ofcircuits when the capacitors were completely charged. But how does the chargebuild up on the capacitor as a function of time? That is, we would like to develop anequation that gives us the charge on a capacitor as a function of time. Let usconsider the simple capacitive circuit shown in figure 25.16, which consists of thecapacitor C, a resistor R1, a battery of emf V, and a switch S. While the switch S is

R

V S

C

Figure 25.16. Charging a capacitor.

open, no charge can flow from the battery to the capacitor. The switch is now closedcompleting the circuit and charge now flows from the battery to the capacitor. If weapply Ohm’s law to the circuit we get

V = VC + VR (25.93)

where VC is the voltage drop across the capacitor, and VR is the voltage drop across


25-41

1 We have indicated a resistor R to be present in the circuit. This is always true, in general, becauseeven if a separate physical resistor is not present, the connecting wires themselves have someresistance.

the resistor. But the voltage drop across the capacitor is given by VC = q/C, and thevoltage drop across the resistor is given by VR = iR, hence

V = q + iR (25.94) C

The current i = dq/dt and replacing this into equation 25.94 we get

V = q + dq R C dt

Upon rearranging− R dq = q − V

dt C

dqdt =

qC − V−R =

q−RC − V

−R

dqdt =

q − VC−RC

dqq − VC = − dt

RCUpon integrating

(25.95)¶0q dq

q − VC = −¶0t dt

RC

Notice that when t = 0, the lower limit on the right-hand side of the equation, thecharge on the capacitor q = 0, the lower limit on the left-hand side of the equation,and when t = t, the upper limit on the right-hand side of the equation, the charge onthe capacitor q = q, the upper limit on the left-hand side of the equation. If we let

v = q − VCthen

dv = dq

and the left-hand side of equation 25.95 is integrated into

¶0q dq

q − VC = ¶ dvv = ln v = ln(q − VC)|0

q = ln(q − VC) − ln(−VC) = ln(q − VC)(−VC)

while the right hand side is integrated into

−¶0t dt

RC = − 1RC t|0

t = − 1RC

(t − 0) = − tRC

Equating the left-hand side to the right-hand side gives


25-42

(25.96)ln(q − VC)(−VC) = − t

RC

Since elnx = x, we now take both sides of equation 25.96 to the e power to get

(q − VC)(−VC) = e

−t

RC

q − VC = −VCe−

tRC

q = VC − VCe−

tRC

q = VC 1 − e−

tRC

But the product VC is equal to the total charge qt on the capacitor when it iscompletely charged. Therefore

(25.97)q = qt 1 − e−

tRC

Equation 25.97 gives the charge q on a charging capacitor at any instant of time t.Notice that when the time t = 0, e0 = 1 and equation 25.97 becomes

q = qt(1− e0) = qt(1 − 1) = 0

The charge on the capacitor at t = 0 is zero because the circuit has just been turnedon and the charge hasn’t had time to get to the capacitor. As t increases, the chargeon the capacitor increases and when t is a very large time, approximated by letting t= ∞, equation 25.97 becomes

q = qt(1 − e−∞ ) = qt 1 − 1e∞ = qt 1 − 1

∞ = qt(1 − 0) = qt

That is, after a large time, the charge on the capacitor is q = qt the total charge. Aplot of equation 25.97 is shown in figure 25.17. Notice that the charge q on thecapacitor starts from the value zero and increases exponentially until it reaches themaximum charge.


25-43

Charge of a capacitor

00.00010.00020.00030.00040.00050.0006

0 0.05 0.1 0.15time (s)

Cha

rge

(C)

Figure 25.17 Charging a capacitor.

Example 25.14

Charging a capacitor. A 5.00 µF capacitor is connected in series with a 5000 Ωresistance to a 100 V battery as shown in figure 25.16. Find (a) the total charge onthe capacitor, and (b) the charge on a capacitor at the time t = 0.045 s.

Solution

a. The total charge on the capacitor when it is completely charged is found as

qt = CV qt = (5.00 × 10−6 F)(100 V)

= 5.00 × 10−4 C

b. The charge on the capacitor at any time is found from equation 25.97 as

q = qt 1 − e−

tRC

q = (5.00 % 10−4 C) 1 − e−

0.045 s(5000 )(5.00 % 10−6 F)

q = (5.00 × 10−4 C)1 − 1.65 × 10−1q = (5.00 × 10−4 C)(8.35 × 10−1)

q = 4.17 × 10−4 C



25-44


An interesting value of the time t is when the time has the numerical valueequal to the product of the resistance and the capacitance, that is for

t = RC (25.98)

This special value of the time is called the time constant of the circuit. Let us findthe amount of charge on the capacitor when t is equal to the time constant.

Example 25.15

Charge on a charging capacitor at the time constant. Find (a) the time constant ofthe circuit in example 25.14, (b) the percentage of the total charge on the capacitorat that time, and (c) the actual value of the charge on the capacitor at that time.

Solution

a. The time constant, found from equation 25.98, is

t = RCt = (5.00 × 103 Ω)(5.00 × 10−6 F) = 2.50 × 10−2 s

b. the percentage of the total charge on the capacitor at that time is found as

q = qt 1 − e−

tRC = qt 1 − e

−RCRC = qt(1 − e−1 )

q = qt1 − 3.68 × 10−1q = 0.632 qt

That is, when t = RC, the time constant of the circuit, the capacitor will be charged to63% of its final charge.

c. the actual value of the charge on the capacitor at that time is found as

q = qt 1 − e−

tRC

q = (5.00 % 10−4 C) 1 − e−

0.0250 s(5000 )(5.00 % 10−6 F)

q = (5.00 × 10−4 C)1 − 0.368q = (5.00 × 10−4 C)(0.632)

q = 3.16 × 10−4 C


25-45


Since the current in the circuit is i = dq/dt, it can be found by differentiatingequation 25.97 as

i =dqdt = qt

ddt 1 − e

−t

RC = qt 0 − ddt e

−t

RC = qt −e−

tRC − 1

RC

i =qt

RC e−

tRC

But qt/C = V the battery voltage, hence

i = VR e

−t

RC

and V/R = I0 the maximum value of the current in the circuit. Therefore the currentbecomes

(25.99)i = I0e−

tRC

Equation 25.99 gives the current i in the circuit as a function of the time t when acapacitor is being charged. Notice that when the time t = 0, e0 = 1 and equation25.99 becomes

i = I0e0 = I0

When the time t = 0, the current in the circuit is at its maximum value I0. This issometime confusing to students because at t = 0 the charge on the capacitor is equalto zero, and yet the current is a maximum. Remember that the current is thederivative of the charge with respect to the time, i.e., i = dq/dt. Even though nocharge has been built up on the capacitor yet, the rate at which the charge isentering the circuit with time is at its maximum value. As t increases the charge qon the capacitor increases, but the current in the circuit decreases from itsmaximum value I0 because as the charge ends up on the plates of the capacitor it isnot available to flow around the circuit as current. When t becomes a very largetime, approximated by letting t = ∞, equation 25.99 becomes

i = I0e−∞ = I0e∞ = I0

∞ = 0

That is, after a large time, the current i in the circuit decreases to zero. The reasonthe current decreases to zero is because all the charge that was originally in thecircuit has all piled up on the plates of the capacitor and is not available to flow inthe circuit as current. A plot of equation 25.99 is shown in figure 25.18. Notice thatthe current i in the capacitor starts from the maximum value I0 and decreases


25-46


Current in the circuit for a charging capacitor

00.0050.01

0.0150.02

0.025

0 0.05 0.1 0.15time (s)

Cur

rent

(A)

Figure 25.18 The current in the circuit while the capacitor is charging.

exponentially until it reaches the value zero. Also notice that figure 25.18 is theopposite of figure 25.17. When t = 0, the charge q in figure 25.17 is equal to zerowhile the current i in figure 25.18 is at its maximum value I0. As the charge q onthe capacitor increases exponentially with time in figure 25.17, the current idecreases exponentially with time in figure 25.18. When t approaches infinity, thecharge q in figure 25.17 increases exponentially approaching its maximum value qt,while the current i in figure 25.18 is decreasing exponentially to its value of zero.

Example 25.16

Current in capacitive circuit of example 25.14 while the capacitor is being charged. A5.00 µF capacitor is connected in series with a 5000 Ω resistance to a 100 V batteryas shown in figure 25.16. Find (a) the maximum value of the current in the circuit,(b) the current in the circuit at the time t = 0.045 s.

Solution

a. The maximum value of the current in the circuit is found from Ohm’s law as

I0 = V/R I0 =(100 V)/(5000 Ω)

= 0.02 A

b. The current in the circuit at any time is found from equation 25.99 as

i = I0e−

tRC

i = (0.02 A)e−

0.045 s(5000 )(5.00 % 10−6 F)


25-47

i = (0.02 A)(0.165)i = 3.30 × 10−3 A


Just as we found the amount of charge on the capacitor at the time equal tothe time constant, let us now find the amount of current in the circuit when t isequal to the time constant.

Example 25.17

The current in a capacitive circuit when the time is equal to the time constant. Find(a) the time constant of the circuit in example 25.14, (b) the percentage of the totalcurrent in the circuit at that time, and (c) the actual value of the current in thecircuit at that time.

Solution


t = RCt = (5.00 × 103 Ω)(5.00 × 10−6 F) = 2.50 × 10−2 s

b. the percentage of the total current in the circuit at that time is found fromequation 25.99 as

i = I0e−

tRC = I0e

−RCRC = I0(e−1 )

i =I00.368i = 0.368 I0

That is, when t = RC, the time constant of the circuit, the current in the chargingcircuit will have decreased to 37% of its initial maximum value. Compare this resultwith example 25.15. There, for the time equal to the time constant, the capacitorwas charged to 63% of its final value, while here we see that the current will havedropped to 37% of its initial value.

c. The actual value of the current i in the circuit for the time t = the time constant isfound from equation 25.99 as

i = I0e−

tRC

i = (0.02 A)e−

0.0250 s(5000 )(5.00 % 10−6 F)


25-48


i = (0.02 A)(0.368)i = 7.36 × 10−3 A


Discharging a CapacitorWe would now like to reverse our process by seeing how the capacitor is

discharged. That is, we would like to develop an equation that gives us the chargeon a capacitor as a function of time when the capacitor is discharging. Let us usethe same capacitive circuit shown in figure 25.16, but we first open the switch S,and remove the battery. The new circuit is shown in figure 25.19. The original

R

S

C

Figure 25.19. Discharging a capacitor.

charge placed on the capacitor is still there. The switch S is now closed completingthe circuit and charge now flows from the capacitor around the circuit through theresistance R. If we apply Ohm’s law to the new circuit, equation 25.93 becomes

0 = VC + VR (25.100)

where the original battery voltage V has now been set equal to zero. VC is still thevoltage drop across the capacitor, and VR is the voltage drop across the resistor. Butthe voltage drop across the capacitor is given by VC = q/C, and the voltage dropacross the resistor is given by VR = iR, hence

0 = q + iR (25.101)C

The current i = dq/dt and replacing this into equation 25.101 we get

0 = q + dq R C dt

Upon rearranging− R dq = q

dt C


25-49


dqq = − dt

RCand integrating

(25.102)¶qt

q dqq = −¶0

t dtRC

Notice that when t = 0, the lower limit on the right-hand side of the equation, thecharge on the capacitor is the total charge q = qt, the lower limit on the left-handside of the equation, and when t = t, the upper limit on the right-hand side of theequation, the charge on the capacitor q = q, the upper limit on the left-hand side ofthe equation. Upon integrating we get

¶qt

q dqq = −¶0

t dtRC

ln q|qtq = − 1

RC t|0t

ln q − ln qt = − 1RC

(t − 0)

(25.103)lnqqt

= − tRC

Since elnx = x, we now take both sides of equation 25.103 to the e power to get

qqt

= e−

tRC

Therefore

(25.104)q = qte−

tRC

Equation 25.104 gives the charge q on a discharging capacitor at any instant of timet. Notice that when the time t = 0, e0 = 1 and equation 25.104 becomes

q = qte0 = qt(1) = qt

The charge on the capacitor at t = 0 is the maximum charge qt because the switchhas just been closed and the charge hasn’t had time to flow into the circuit. As tincreases the charge on the capacitor decreases and when t is a very large time,approximated by letting t = ∞, equation 25.104 becomes

q = qte−∞ =qte∞ =

qt∞ = 0

That is, after a large time, the charge on the capacitor is q = 0, because thecapacitor has been completely discharged. A plot of equation 25.104 is shown infigure 25.20. Notice that the charge q on the capacitor starts from the maximumvalue qt and decreases exponentially until it reaches the value zero.


25-50

Discharge of a capacitor

00.00010.00020.00030.00040.00050.0006

0 0.05 0.1 0.15time (s)

Cha

rge

(C)

Figure 25.20 Discharging a capacitor.

Example 25.18

Discharging a capacitor. A 5.00 µF capacitor carries a charge qt = 5.00 × 10−4 C. It isconnected in series with a 5000 Ω resistance as shown in figure 25.19. Find thecharge on the capacitor at the time t = 0.045 s.

Solution

The charge on the capacitor at any time is found from equation 25.104 as

q = qte−

tRC

q = (5.00 % 10−4 C) e−

0.045 s(5000 )(5.00 % 10−6 F)

q = (5.00 × 10−4 C)(1.65 × 10−1)q = 8.25 × 10−5 C


Let us now find the amount of charge left on the capacitor when t is equal tothe time constant.

Example 25.19

Charge on a discharging capacitor at the time constant. Find (a) the time constant of


25-51

http://www.farmingdale.edu/faculty/peter-nolan/pdf/univphys/UPEX25-18.xls

the circuit in example 25.18, (b) the percentage of the total charge left on thecapacitor at that time, and (c) the actual value of the charge on the capacitor at thattime.

Solution


t = RCt = (5.00 × 103 Ω)(5.00 × 10−6 F) = 2.50 × 10−2 s

b. The percentage of the total charge left on the capacitor at that time is found as

q = qt e−

tRC = qt e

−RCRC = qt(e−1 )

q = qt( 3.68 × 10−1)q = 0.368 qt

That is, when t = RC, the time constant of the circuit, the charge left on thedischarging capacitor will be 37% of its initial charge. Compare this result withexample 25.15 for a charging capacitor. At the time of the time constant for acharging capacitor, the capacitor was 63% fully charged, while for the dischargingcapacitor the capacitor is only 37% full for the same time constant. Or another wayto look at it, it gained 63% of its total charge during the time constant while it wascharging and lost 63% of its total charge during the same time constant when it wasdischarging.

c. the actual value of the charge on the capacitor at that time is found as

q = qte−

tRC

q = (5.00 % 10−4 C) e−

0.0250 s(5000 )(5.00 % 10−6 F)

q = (5.00 × 10−4 C)(0.368)q = 1.84 × 10−4 C


Let us now find the current in the discharging capacitor. Since the current inthe circuit is i = dq/dt, it can be found by differentiating equation 25.104 as


25-52


i =dqdt = qt

ddt e

−t

RC = qt e−

tRC − 1

RC

i = −qt

RC e−

tRC

But qt/C = V the initial voltage across the capacitor, hence

i = −VR e

−t

RC

and V/R = I0 the maximum value of the current in the circuit. Therefore the currentbecomes

(25.104)i = −I0e−

tRC

Equation 25.104 gives the current i in the circuit as a function of the time t when acapacitor is being discharged. Notice that this equation gives the same currentwhen the capacitor was charging except for the minus sign. The minus sign saysthat the current in now flowing in the opposite direction from when the capacitorwas charging. When the time t = 0, e0 = 1 and equation 25.104 becomes

i = −I0e0 = −I0

When the time t = 0, the current in the circuit is at its minimum value −I0. Thecurrent is negative because the charge is now coming out of the capacitor ratherthan going in. Remember that the since the current is the derivative of the chargewith respect to the time, i.e., i = dq/dt, the rate at which the charge is leaving thecapacitor with time is at its maximum value. As t increases the charge q on thecapacitor decreases, and the current in the circuit increases from its minimum value−I0. When t becomes a very large time, approximated by letting t = ∞, equation25.104 becomes

i = −I0e−∞ = − I0e∞ = −I0

∞ = 0

That is, after a large time, the current i in the circuit increases from −I0 to zero. Thereason the current increases to zero is because all the positive charge that wasoriginally on the positive plate of the capacitor has moved to the negative plate ofthe capacitor, neutralizing all the negative charges that were there. Hence there isno more free charge to move around the circuit, and the current is zero. A plot ofequation 25.104 is shown in figure 25.21. Notice that the current i in the circuitstarts from the minimum value −I0 and increases exponentially until it reaches thevalue zero. Also notice that figure 25.21 is the opposite of figure 25.20. When t = 0,the charge q in figure 25.20 is equal to the maximum value qt while the current i infigure 25.21 is at its minimum value −I0. As the charge q on the capacitor decreases


25-53

exponentially with time in figure 25.20, the current i increases exponentially withtime in figure 25.21. When t approaches infinity, the charge q in figure 25.20decreases exponentially approaching its minimum value zero, while the current i infigure 25.21 is increasing exponentially to its maximum value of zero.

Current in the circuit for a discharging capacitor

-0.025-0.02

-0.015-0.01

-0.0050

0 0.05 0.1 0.15

time (s)

Cur

rent

(A)

Figure 25.21 The current in the circuit while capacitor is discharging.

Example 25.20

Current in capacitive circuit of example 25.18 while the capacitor is beingdischarged. A 5.00 µF capacitor, which has a 100 V potential drop across it, isconnected in series with a 5000 Ω resistance as shown in figure 25.19. Find (a) themaximum value of the current in the circuit, (b) the current in the circuit at thetime t = 0.045 s.

Solution

a. The maximum value of the current in the circuit is found from Ohm’s law as

I0 = V/R I0 =(100 V)/(5000 Ω)

= 0.02 A

b. The current in the circuit at any time is found from equation 25.104 as

i = −I0e−

tRC

i = −(0.02 A)e−

0.045 s(5000 )(5.00 % 10−6 F)

i = −(0.02 A)(0.165)i = − 3.30 × 10−3 A


25-54


Let us now find the amount of current in the circuit of a discharging capacitorwhen t is equal to the time constant.

Example 25.21

The current in a discharging capacitive circuit when the time is equal to the timeconstant. Find (a) the time constant of the circuit in example 25.20, (b) thepercentage of the total current in the circuit at that time, and (c) the actual value ofthe current in the circuit at that time.

Solution


t = RCt = (5.00 × 103 Ω)(5.00 × 10−6 F) = 2.50 × 10−2 s

b. the percentage of the total current in the circuit at that time is found fromequation 25.104 as

i = −I0e−

tRC = −I0e

−RCRC = −I0(e−1 )

i = −I0(0.368)i = −0.368 I0

That is, when t = RC, the time constant of the circuit, the current in the dischargingcircuit will have decreased to 37% of its initial minimum value.

b. The actual value of the current i in the circuit for the time t = the time constantis found from equation 25.104 as

i = −I0e−

tRC

i = −(0.02 A)e−

0.0250 s(5000 )(5.00 % 10−6 F)

i = −(0.02 A)(0.368)i = − 7.36 × 10−3 A



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The Language of Physics

CapacitorTwo conductors of any size or shape carrying equal and opposite charges is called acapacitor. The charge on the capacitor is directly proportional to the potentialdifference between the plates. The importance of the capacitor lies in the fact energycan be stored in the electric field between the two bodies (p. ).

Energy DensityThe energy per unit volume that is stored in the electric field (p. ).

Capacitors in seriesWhen capacitors are connected in series, each capacitor has the same charge on itsplates. The reciprocal of the equivalent capacitance is equal to the sum of thereciprocals of each capacitor (p. ).

Capacitors in parallelWhen capacitors are connected in parallel there is a different amount of chargedeposited on the plates of each capacitor, but the potential difference is the sameacross each of the parallel capacitors. The equivalent capacitance is equal to thesum of the individual capacitances (p. ).

Dielectric constantThe dielectric constant is defined as the ratio of the capacitance with a dielectricbetween the plates to the capacitance with air or vacuum between the plates (p. ).

Capacitors with a dielectricPlacing a dielectric, an insulator, between the plates of a capacitor increases thecapacitance of that capacitor; decreases the electric field and the potential differencebetween the plates; decreases the amount of energy that can be stored in thecapacitor; and increases the dielectric strength of the capacitor (p. ).

Dielectric strengthThe value of the potential difference per unit plate separation when the dielectricbreaks down (p. ).

Time constantThe value of the time t in a capacitive circuit when the time has the numericalvalue equal to the product of the resistance and the capacitance, that is for t = RC.For a charging capacitor, when t = RC, the capacitor will be charged to 63% of itsfinal charge. For a discharging capacitor, when t = RC, the capacitor will be chargedto 37% of its initial charge (p. ).


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Summary of Important Equations

Charge on a Capacitor q = CV (25.15)

Definition of Capacitance C = q/V (25.16)

Parallel plate capacitorPotential between the plates V = Ed (25.3)Electric field between the plates (25.11)E =

qoA

Capacitance (25.14)C = oAd Coaxial cylindrical capacitorElectric field between the cylinders (25.29)E =

q2orl

Potential between the cylinders (25.30)V =q

20l ln rbra

Capacitance (25.32)C = 2olln(rb/ra )

Concentric spherical capacitorElectric field between the spheres (25.40)E = 1

4oqr2 =

kqr2

Potential between the spheres (25.41)V = kq( rb − rararb

)

Capacitance (25.43)C = 4orarbrb − ra

Energy stored in a Capacitor W = 1/2 qV (25.52) W = 1/2 CV2 (25.50)

W = 1/2 q2/C (25.48)

Energy density in electric field UE = 1/2 εoE2 (25.57) Equivalent capacitance of capacitors in series

1 = 1 + 1 + 1 + … (25.64) C C1 C2 C3

Equivalent capacitance of capacitors in parallel C = C1 + C2 + C3 + … (25.68)

Definition of dielectric constant κ = Cd (25.74) C0

Capacitance of capacitor with a dielectric Cd = κC0 (25.75)


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Permittivity of the dielectric medium ε = κε0 (25.77)

Charge on a charging capacitor at any instant of time (25.97)q = qt 1 − e−

tRC

Time constant t = RC (25.98)

Current in a circuit as a function of time when capacitor is being charged

(25.99)i = I0e−

tRC

Charge on a discharging capacitor at any instant of time

(25.104)q = qte−

tRC

Current in a circuit as a function of time when capacitor is being discharged

(25.104)i = −I0e−

tRC

Questions for Chapter 251. Can you speak of the capacitance of an isolated conducting sphere?2. How does the direction signal device on your car make use of a capacitor?3. Discuss the statement, “If a capacitor contains equal amounts of positive

and negative charge it should be neutral and have no effect whatsoever.’’4. If the potential difference across a capacitor is doubled what does this do to

the energy that the capacitor can store?5. If an ammeter is connected in series to the circuit of figure 25.2(a), what

will it indicate? Why?6. In tuning in your favorite radio station, you adjust the tuning knob, which

is connected to a variable capacitor. How does this variable capacitor work?7. Is more energy stored in capacitors when they are connected in parallel or

in series? Why?8. If the applied potential difference across a capacitor exceeds the dielectric

strength of the medium between the plates, what happens to the capacitor?9. Can a coaxial cable have a capacitance associated with it?

Problems for Chapter 2525.2 The Parallel Plate Capacitor

1. Find the charge on a 4.00-µF capacitor if it is connected to a 12.0-Vbattery.

2. If the charge on a 9.00-µF capacitor is 5.00 × 10−4 C, what is the potentialacross it?

3. How much charge must be removed from a capacitor such that the newpotential across the plates is 1/2 of the original potential?


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4. A charge of 6.00 × 10−4 C is found on a capacitor when a potential of 120 Vis placed across it. What is the value of the capacitance of this capacitor?

5. A parallel plate capacitor has plates that are 6.00 cm by 4.00 cm and areseparated by 8.00 mm. Find the capacitance of the capacitor.

6. If a 5.00-µF parallel plate capacitor has its plate separation doubled whileits cross-sectional area is tripled, determine the new value of its capacitance.

7. You are asked to design your own parallel plate capacitor that is capable ofholding a charge of 3.60 × 10−12 C when placed across a potential difference of 12.0V. If the area of the plates is equal to 100 cm2 what must the plate separation be?What is the value of the capacitance of this capacitor?

8. You are asked to design a parallel plate capacitor that can hold a charge of9.87 pC when a potential difference of 24.0 V is applied across the plates. Find theratio of the area of the plates to the plate separation, and then pick a reasonable setof values for them.

25.3 The cylindrical Capacitor9. A cylindrical capacitance 10.0 cm long has an inner radius of 0.500 mm

and an outside radius of 5.00 mm. Find the capacitance of the capacitor.10. A cylindrical capacitor 1.00 m long has radii 20.0 cm and 50.0 cm. Find its

capacitance.

25.4 The spherical Capacitor11. A spherical capacitor has radii 20.0 cm and 50.0 cm. Find its capacitance.12. Show that if the radii of a spherical capacitor are very large, the equation

for the capacitance of a spherical capacitor reduces to the equation for a parallelplate capacitor.

13. Find the formula for the capacitance of an isolated sphere. (Hint: use therelation for a concentric spherical capacitor and let the radius of the outer sphere goto infinity.)

14. Consider the earth to be an isolated sphere. Find the capacitance of theearth. If the electric field of the earth is measured to be 100 V/m, what charge mustbe on the surface of the earth? (Note that the fair weather electric field pointsdownward, indicating an effective negative charge on the solid sphere.)

25.5 Energy Stored in a Capacitor15. A 7.00-µF capacitor is connected to a 400-V source. Find (a) the charge on

the capacitor and (b) the energy stored in the capacitor.16. How much energy can be stored in a capacitor of 9.45 µF when it is placed

across a potential difference of 120 V? How much charge will be on this capacitor?17. What value of capacitance is necessary to store 1.73 × 10−3 J of energy

when it is placed across a potential difference of 24.0 V?


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25.6 Capacitors in Series18. Find the equivalent capacitance of 2.00-µF, 4.00-µF, and 8.00-µF

capacitors connected in series.19. Find (a) the equivalent capacitance, (b) the charge on each capacitor,

(c) the voltage across each capacitor, and (d) the energy stored in each capacitor inthe diagram.

Diagram for problem 19.

25.7 Capacitors in Parallel20. Find the equivalent capacitance of 2.00-µF, 4.00-µF, and 8.00-µF

capacitors connected in parallel.21. Find the charge on each capacitor and the energy stored in each capacitor

in the diagram.

Diagram for problem 21. Diagram for problem 22.

22. A 3.00-µF capacitor is initially connected to a battery and charged to 100V. It is then removed and connected in parallel to a 15.0-µF capacitor that wasuncharged. Find (a) the initial charge on the first capacitor, (b) the charge on eachcapacitor after being connected in parallel, (c) the initial energy stored in the firstcapacitor, and (d) the energy stored in each capacitor after they are connected inparallel.

23. A 40.0-µF capacitor and an 80.0-µF capacitor are initially connected inparallel across a 10.0-V potential difference. The two capacitors are thendisconnected from each other and from the potential source, and reconnected with


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plates of unlike sign connected together. Find the charge on each capacitor and thepotential difference across each capacitor after the two capacitors are so connected.

25.8 Combinations of Capacitors in Series and Parallel24. Find (a) the equivalent capacitance of the circuit, (b) the total charge

drawn from the battery, (c) the voltage across each capacitor, (d) the charge on eachcapacitor, and (e) the energy stored in each capacitor in the accompanying diagram.


25. Find the equivalent capacitance of the circuit diagram if C1 = 1.00 µF, C2 =2.00 µF, C3 = 3.00 µF, C4 = 4.00 µF, and C5 = 5.00 µF.

26. In the accompanying diagram find (a) the equivalent capacitance, (b) thecharge on capacitor C2, and (c) the voltage drop across C4, if C1 = 10.0 µF, C2 = 20.0µF, C3 = 30.0 µF, C4 = 40.0 µF, and E = 12.0 V.


27. The network of capacitors in the diagram is connected across a potentialdifference V = 3.00 V. If C1 = C2 = 30.0 µF, C3 = 60.0 µF, and C4 = C5 = C6 = 20.0 µF,find (a) the equivalent capacitance of the network, (b) the charge on each capacitor,(c) the potential across each capacitor, and (d) the energy stored in each capacitor.

28. Find the equivalent capacitance and the charge on each capacitor in thediagram if C1 = 5.00 µF, C2 = 15.00 µF, C3 = 8.00 µF, C4 = 9.00 µF, and E = 12.0 V


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25.9 and 25.10 Capacitors with Dielectrics Placed between the Plates29. A 5.00-µF parallel plate capacitor has air between the plates. When an

insulating material is placed between the plates, the capacitance increases to 13.5µF. Find the dielectric constant of the insulator.

30. If the dielectric constant of rubber is 2.94, what is its permittivity?31. Find the capacitance of a parallel plate capacitor having an area of 75.0

cm2, a plate separation of 3.00 mm, and a strip of bakelite placed between theplates.

32. A 350-µF capacitor is constructed with bakelite between its plates. If thebakelite is removed and replaced by amber, find the new capacitance.

33. A 0.100-mm piece of wax paper is to be placed between two pieces ofaluminum foil to make a parallel plate capacitor of 5 µF. What must the area of thefoil be?

34. A parallel plate capacitor has plates that are 5.00 cm by 6.00 cm and areseparated by an air gap of 1.50 mm. Calculate the maximum voltage that can beapplied to the capacitor before dielectric breakdown.

35. A 1.00-mm piece of wax paper is placed between two pieces of aluminumfoil that are 10.0 cm by 10.0 cm. How much energy can be stored in this capacitor ifit is connected to a 24.0 V battery?

36. If 1.73 × 10−3 J of energy can be stored in a capacitor with air between theplates, how much energy can be stored in the capacitor if a slab of mica is placedbetween the plates?

37. What is the maximum charge that can be placed on the plates of an aircapacitor of 9.00 µF before breakdown if the plate separation is 1.00 mm? If a stripof mica is placed between the plates, what is the maximum charge?

38. A parallel plate capacitor has a value of 3.00 pF. A 1.00-mm piece ofplexiglass is placed between the plates of the capacitor, which is then connected to a12.0-V battery. Find (a) the electric field E0 without the dielectric between theplates and (b) the induced electric field in the dielectric Ei.

39. A sample of NaCl is placed into a solution of ethyl alcohol. If the distancebetween the Na+ and the Cl− ions is 1.00 × 10−7 cm, find the force on the ions.


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25.11 Charging and Discharging a Capacitor40. A 7.50 µF capacitor is connected in series with a 2500 Ω resistance to a 12

V battery as shown in figure 25.16. Find (a) the total charge on the capacitor, (b) thecharge on a capacitor at the time t = 0.015 s, (c) the time constant of the circuit, (d)the percentage of the total charge on the capacitor at that time, and (e) the actualvalue of the charge on the capacitor at that time.

41. A 3.25 µF capacitor is connected in series with a 4000 Ω resistance to a15.5 V battery as shown in figure 25.16. Find (a) the maximum value of the currentin the circuit, (b) the current in the circuit at the time t = 0.035 s, (c) the timeconstant of the circuit, (d) the percentage of the total current in the circuit at thattime, and (e) the actual value of the current in the circuit at that time.

42. A 4.25 µF capacitor carries a charge qt = 7.52 × 10−4 C. It is connected inseries with a 4500 Ω resistance as shown in figure 25.19. Find (a) the charge on thecapacitor at the time t = 0.02 s, (b) the time constant of the circuit, (c) thepercentage of the total charge left on the capacitor at that time, and (d) the actualvalue of the charge on the capacitor at that time.

43. A 9.75 µF capacitor, which has a 50.0 V potential drop across it, isconnected in series with a 7500 Ω resistance as shown in figure 25.19. Find (a) themaximum value of the current in the circuit, (b) the current in the circuit at thetime t = 0.00525 s, (c) the time constant of the circuit, (d) the percentage of the totalcurrent in the circuit at that time, and (e) the actual value of the current in thecircuit at that time.

Additional Problems44. Determine all the values of capacitance that you can obtain from the

three capacitors of 3.00 µF, 6.00 µF, and 9.00 µF.45. (a) Find the equivalent capacitance of 50 identical 2.50-µF capacitors in

series. (b) Find the equivalent capacitance of 50 identical 2.50-µF capacitors inparallel.

46. A potential of 24.0 V is applied across capacitor C1 = 2.50 µF. Find thecharge on capacitors C2 and C3 if C2 = 6.00 µF and C3 = 4.00 µF.



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47. In the accompanying diagram find (a) the equivalent capacitance, (b) thevoltage drop across each capacitor, and (c) the charge on each capacitor if C1 = 3.00µF, C2 = 6.00 µF, C3 = 9.00 µF, C4 = 12.00 µF, C5 = 15.00 µF, C6 = 18.00 µF, and E 1 =12.0 V.

48. Find the potential drop across AB in the diagram if C1 = 3.00 µF, C2 =6.00 µF, C3 = 9.00 µF, C4 = 12.00 µF, and E = 24.0 V.


49. In the accompanying circuit, find (a) the final value of current recorded byan ammeter that is first placed in series with the resistor R, then the capacitor C1,and finally the capacitor C2; (b) find the voltage drop recorded by a voltmeter whenplaced across the resistor R, the capacitor C1, and the capacitor C2; and (c) find thecharge on C1 and C2.

50. A cylindrical capacitor is made by wrapping a sheet of aluminum foilaround a hollow cardboard core 2.50 cm in diameter. A 1.00 mm thick piece of waxpaper is then wrapped around the foil and then another sheet of aluminum foil iswrapped around the wax paper. The length of the core is 25.0 cm. Find thecapacitance of this cylindrical capacitor.

51. Find (a) the capacitance of a parallel plate capacitor having an area of40.0 cm2 and separated by air 2.00 mm thick, (b) the capacitance if 2.00 mm of micais placed between the plates, (c) if the capacitor is connected to a 6.00-V battery,find the charge on the capacitor with air and with mica between the plates, (d) theelectric field between the plates, and (e) the energy stored in each capacitor.

52. The electric field between the plates of a parallel plate capacitor filledwith air is E0 = q/(ε0A). When a dielectric is placed between the plates, the inducedelectric field within the dielectric is given by Ei = q’ /(ε0A), where q’ is the boundcharge. (a) Show that the bound charge on the dielectric is given by

q’ = q(1 − 1 ) κ

(b) If a piece of mica is placed between the plates of a capacitor of 6.00 µF and isthen connected to the plates of a 24.0-V battery, find the charge q and q’.


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53. Two identical parallel plate capacitors, one with air between the plates,and the other with mica, are connected to a 100-V battery, as shown in the diagram.If C1 = 3.00 µF and the plate separation is 0.500 mm, find (a) the charge on eachplate, (b) the electric field between the plates, and (c) the energy stored in eachcapacitor.


54. A cumulonimbus cloud is 5.00 km long by 5.00 km wide and has its base1.00 km above the surface of the earth, as shown in the diagram. Consider the cloudand earth to be a parallel plate capacitor with air as the dielectric. (a) Find thecapacitance of the cloud-earth combination. (b) Find the potential differencebetween the cloud and the earth when lightning occurs. (c) Calculate the chargethat must be on the base of the cloud when lightning occurs.

55. A parallel plate capacitor with air between the plates has a plateseparation d0 and area A. A sheet of aluminum of negligible thickness is now placedmidway between the parallel plates. Find the capacitance (a) before and (b) afterthe aluminum sheet is placed between the plates.

56. A parallel plate capacitor with air between the plates has a plateseparation d0 and area A. A half of a sheet of aluminum of negligible thickness isnow placed midway between the parallel plates as shown. Show what thiscombination is equivalent to and find the equivalent capacitance.



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57. A parallel plate capacitor with dielectric constant κ1 is connected in serieswith a similar parallel plate capacitor except that it has a dielectric constant κ2

between the plates, as shown in the diagram. (a) Find a single equation for theequivalent capacitor. (b) What is the value of an equivalent dielectric that could beplaced in one of the original capacitors to give an equivalent capacitor?

58. A parallel plate capacitor has a moveable top plate that executes simpleharmonic motion at a frequency f. The displacement of the top plate varies suchthat the plate separation varies from d to 2d. Show that the voltage across MNvaries as

V = V0(1 + 1 cos 2πft) 2

where V0 is the voltage across the capacitor when the plates are at a constantseparation d.


59. One model of a red blood cell considers the cell to be a spherical capacitor.The radius of a blood cell is about 6.00 × 10−4 cm and the membrane wall of 0.100 ×10−4 cm is considered to be a dielectric material with a dielectric constant of 5.00.Find (a) the capacitance of the cell. (b) If a potential difference of 100 mV ismeasured across the blood cell, find the charge on a red blood cell.

60. When you tune in your favorite radio station, you use a variable capacitorin the radio. A variable capacitor consists of a series of ganged metal plates, asshown in the diagram. By rotating the plates, the effective area of the capacitor canbe changed, thus changing the capacitance of the capacitor, which in turn changesthe frequency of the tuned circuit. Show that the equivalent capacitance of thecapacitor shown is

C = 5ε0A d


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Interactive Tutorials61. A parallel plate capacitor. A parallel plate capacitor has plates that have

an area A = 83.5 cm2 and a plate separation d = 6.00 mm. The space between theplates is filled with air. The capacitor is connected to a 12.0-V battery. Find (a) thecapacitance C of the capacitor, (b) the charge q deposited on its plates, and (c) theenergy stored in the capacitor.

62. Capacitors in series. Three capacitors, C1 = 2.55 × 10−6 F, C2 = 5.35 × 10−6

F, and C3 = 8.55 × 10−6 F are connected in series to a 100-V battery. Find (a) theequivalent capacitance, (b) the charge on each capacitor, (c) the voltage drop acrosseach capacitor, (d) the energy stored in each capacitor, and (e) the total energystored in the circuit.

63. Capacitors in parallel. Three capacitors, C1 = 2.55 × 10−6 F, C2 = 5.35 ×10−6 F, and C3 = 8.55 × 10−6 F are connected in parallel to a 100-V battery. Find (a)the equivalent capacitance, (b) the total charge on the equivalent capacitor, (c) thevoltage drop across each capacitor, (d) the charge on each capacitor, (e) the energystored in each capacitor, and (f) the total energy stored in the circuit.

64. Combination of capacitors in series and parallel. Capacitor C1 = 2.55 ×10−6 F, is in series with capacitors C2 = 5.35 × 10−6 F and C3 = 8.55 × 10−6 C, whichare in parallel with each other. The entire combination is connected to a 100-Vbattery. A similar schematic is shown in figure 23.6. Find (a) the equivalentcapacitance of the combination, (b) The total charge q on the equivalent capacitor,(c) the charge q1 on capacitor C1, (d) the voltage drop across each capacitor, (e) thecharge on capacitors C2 and C3, (f) the energy stored in each capacitor, and (g) thetotal energy stored in the circuit.

65. A capacitor with a dielectric. A parallel plate capacitor, having a platearea A = 3.8 × 10−3 m2 and a plate separation d = 1.75 mm, is charged by a batteryto Vo = 85.0 V. The battery is then removed. Find (a) the capacitance of thecapacitor and (b) the charge on the plates of the capacitor. A slab of mica is thenplaced between the plates of the capacitor. Find (c) the new value of the capacitance,(d) the potential difference across the capacitor with the dielectric, (e) the initialelectric field between the plates, (f) the final electric field between the plates, (g) theinitial energy stored in the capacitor, and (h) the final energy stored in thecapacitor.


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66. Charging a capacitor. A 7.50 µF capacitor is connected in series with a5500 Ω resistance to a 110 V battery as shown in figure 25.16. Find (a) the totalcharge on the capacitor, and (b) the charge on a capacitor at the time t = 0.015 s. (c)the time constant of the circuit, and (d) the percentage of the total charge on thecapacitor at that time. (e) the actual charge at that time, and (f) Plot the charge onthe capacitor as a function of time.

67. Current in a circuit while the capacitor is charging. An 8.75 µF capacitoris connected in series with a 2550 Ω resistance to a 50.5 V battery as shown infigure 25.16. Find (a) the maximum value of the current in the circuit, (b) thecurrent in the circuit at the time t = 0.035 s. (c) Plot the current in the circuit as afunction of time.

68. Discharging a capacitor. A 4.25 µF capacitor carries a charge qt = 7.52 ×10−4 C. It is connected in series with a 4500 Ω resistance as shown in figure 25.19.Find (a) the charge on the capacitor at the time t = 0.028 s. (b) the time constant ofthe circuit (c) the percentage of the total charge on the capacitor at that time. (d) theactual charge at that time, and (e) Plot the charge on the capacitor as a function oftime.

69. Current in a circuit while the capacitor is discharging. A 3.65 µF capacitorthat is discharging, has an initial current I0 = 0.035 A, is connected in series with a6500 Ω resistance as shown in figure 25.19. Find (a) the current in the circuit at thetime t = 0.0185 s. (b) the time constant of the circuit (c) the percentage of the initialcurrent in the circuit at that time. (d) the actual current at that time, and (e) Plotthe current in the circuit as a function of time.

To go to these Interactive Tutorials click on this sentence.

To go to another chapter, return to the table of contents byclicking on this sentence.


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