1. 2 Today’s agendum: Electric potential energy. You must be able to use electric potential energy...

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Today’s agendum:

Electric potential energy.You must be able to use electric potential energy in work-energy calculations.

Electric potential.You must be able to calculate the electric potential for a point charge, and use the electric potential in work-energy calculations.

Electric potential and electric potential energy of a system of charges.You must be able to calculate both electric potential and electric potential energy for a system of charged particles (point charges today, charge distributions next lecture).

The electron volt.You must be able to use the electron volt as an alternative unit of energy.

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This lecture introduces electric potential energy and something called “electric potential.”

Electric potential energy is “just like” gravitational potential energy.

Except that all matter exerts an attractive gravitational force, but charged particles exert either attractive or repulsive electrical forces—so we need to be careful with our signs.

Electric potential is the electric potential energy per unit of charge.

If you understand the symbols in the starting equations, and avoid sign and direction mistakes, homework and exams are not difficult.

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Electric PotentialElectric Potential Energy

Electric Potential Energy

b

aFE

r

drds

q2 (-)

q1 (+)

Work done by Coulomb force when q1 moves from a to b:

b b

a a

r r 1 2E 2r r

k q qW F ds dr

r

bb

aa

rr

1 2 1 22rr

1 1W k q q dr k q q

r r

1 2 1 2b a b a

1 1 1 1W k q q k q q

r r r r

ra

rb

q1 (+)

You don’t need to worry about the details of the math. They are provided for anybody who wants to study them later.

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1 2b a

1 1W k q q

r r

I did the calculation for a + charge moving away from a – charge; you could do a similar calculation for ++, -+, and ++.

The important point is that the work depends only on the initial and final positions of q1.

In other words, the work done by the electric force is independent of path taken. The electric force is a conservative force.

b

aFE

r

drds

q2 (-)

q1 (+)

ra

rb

Disclaimer: this is a “demonstration” rather than a rigorous proof.

6

The next two slides are intended to draw a parallel between the electric and gravitational forces.

Instead of saying “ugh, this is confusing new stuff,” you are supposed to say, “oh, this is easy, because I already learned the concepts in Physics 103.”

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If released, it gains kinetic energy and loses potential energy, but mechanical energy is conserved: Ef=Ei. The change in potential energy is Uf - Ui = -(Wc)if.

A bit of review:

y

graphic “borrowed” from http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html

What force does Wc? Force due to gravity.

x

Ui = mgyi

Uf = 0

yi

Consider an object of mass m in a gravitational field. It has potential energy U(y) = mgy and “feels” a gravitational force FG = GmM/r2, attractive.

http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html

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+ + + + + + + + + + + + + +

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

+

E

A charged particle in an electric field has electric potential energy.

It “feels” a force (as given by Coulomb’s law).

It gains kinetic energy and loses potential energy if released. The Coulomb force does positive work, and mechanical energy is conserved.

F

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The next two slides define electrical potential energy.

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? on FE means the direction depends on the signs of the charges.

Now that we realize the electric force is conservative, we can define a potential energy associated with it.

E Ef Ei E i fU U U W

The change in potential energy when a charge q0 moves from point a to point b in the electric field of another charge q is

b

aFE?

r

drds

q

q0

ra

rab

b b

a a

r r0

E E 2r r

k qqU F d dr

r

The minus sign in this equation comes from the definition of change in potential energy. The sign from the dot product is “automatically” correct if you include the signs of q and q0.

The subscript “E” is to remind you this is electric potential energy. After this slide, I will drop the subscript “E.”

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The next two slides use this definition of electrical potential energy to derive an equation for the electrical potential energy of two charged particles.

is equivalent to your starting equation

b

a

r

ErU F d

f

f i iU U q E d

“i” and “f” refer to the two points for which we are calculating the potential energy difference. You could also use “a” and “b” like your text does, or “0” and “1” or anything else convenient. I use “i” and “f” because I always remember that (anything) = (anything)f – (anything)i.

(from the previous slide)

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b

aFE?

r

drds

q

q0

ra

rab

bb

aa

rr

b a 0 02rr

1 1U U k qq dr k qq

r r

b a 0b a

1 1U U k qq

r r

By convention, we choose electric potential energy to be zero at infinite separation of the charges.

b 0b

1 1U U k qq

r

0 0

If there are any math majors in the room, please close your eyes for a few seconds. We should be talking about limits.

starting with an equation from two slides back…

this diagram shows q0 after it has moved from a to b

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This provides us with the electric potential energy for a system of two point charges q and q0, separated by a distance r:

00

0

qq1 1U r k qq .

r 4 r

Example: calculate the electric potential energy for two protons separated by a typical proton-proton intranuclear distance of 2x10-15 m.

What is the meaning of the + sign in the result?

You can call the charges q and q0, or q1 and q2, or whatever you want.If you have more than two charged particles, simply add the potential energies for each unique pair of

particles.

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Really Important fact to keep straight.

f i conservative i fU U U W

The change in potential energy is the negative of the work done by the conservative force which is associated with the potential energy (the electric force).

If an external force* moves an object “against” the conservative force, then

external conservativei f i fW W

Always ask yourself which work you are calculating.

*for example, if you push two negatively charged balloons together

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Today’s agendum:





16

Electric Potential

Previously, we defined the electric field by the force it exerts on a test charge q0:

0

0

q 00

FE = lim

q

Similarly, it is useful to define the potential of a charge in terms of the potential energy of a test charge q0:

0q 0

0

U rV r = lim

q

The electric potential V is independent of the test charge q0.

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From

we see that the electric potential of a point charge q is

0

0

qq1U r

4 r

0

1 qV r .

4 r

The electric potential difference between points a and b is b

b ba

a a

r

E r rr E

r r0 0 0

F d FUV d E d .

q q q

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Things to remember about electric potential:

Electric potential and electric potential energy are related, but are not the same.

Electric potential difference is the work per unit of charge that must be done to move a charge from one point to another without changing its kinetic energy.

The terms “electric potential” and “potential” are used interchangeably.

0

U rV r = .

q

The units of potential are joules/coulomb:

1 joule1 volt =

1 coulomb

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Things to remember about electric potential:

Only differences in electric potential and electric potential energy are meaningful.

It is always necessary to define where U and V are zero. Here we defined V to be zero at an infinite distance from the sources of the electric field.

Sometimes it is convenient to define V to be zero at the earth (ground).

It should be clear from the context where V is defined to be zero, and I do not foresee you experiencing any confusion about where V is zero.

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Two more starting equations:

0

1 qV r

4 r

0

0

qq1U r

4 r

and

so 0

0 0 0 0

U r qq1 1 1 qV(r)

q 4 r q 4 r

(potential is equal to potential energy per unit of charge)

Potential energy and electric potential are defined relative to some reference point, so it is “better” to use

f i

UV V V

q

21

f

f i iU U q E d

ff i

i

U UE d

q q

f

f i iV V E d

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Two conceptual examples.

Example: a proton is released in a region in space where there is an electric potential. Describe the subsequent motion of the proton.

Example: an electron is released in a region in space where there is an electric potential. Describe the subsequent motion of the electron.

The proton will move towards the region of lower potential. As it moves, its potential energy will decrease, and its kinetic energy and speed will increase.

The electron will move towards the region of higher potential. As it moves, its potential energy will decrease, and its kinetic energy and speed will increase.

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Today’s agendum:





24

Electric Potential Energy of a System of Charges

To find the electric potential energy for a system of two charges, we bring a second charge in from an infinite distance away:

before

after

q1

U 0

q1 q2

1 2q qU k

r

r

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To find the electric potential energy for a system of three charges, we bring a third charge in from an infinite distance away:

before

q1

after

q1 q2

1 3 2 31 2

12 13 23

q q q qq qU k

r r r

r12

q2

r12

1 2

12

q qU k

r

q3

r13 r23

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Electric Potential and Potential Energy of a Charge Distribution

Collection of charges: iP

i0 i

q1V .

4 r

Charge distribution:

P is the point at which V is to be calculated, and ri is the distance of the ith charge from P.

Pr

dq

0

1 dqV .

4 r

Potential at point P.

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Example: a 1 C point charge is located at the origin and a -4 C point charge 4 meters along the +x axis. Calculate the electric potential at a point P, 3 meters along the +y axis.

q2q1

3 m

P

4 mx

yi 1 2

Pi i 1 2

-6 -69

3

q q qV = k = k +

r r r

1×10 -4×10= 9×10 +

3 5

= - 4.2×10 V

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Example: how much work is required to bring a +3 C point charge from infinity to point P? An external force moves q3 slowly from an infinite distance to the point P.

q2q1

3 m

P

4 mx

y

q3

external 3W U q V

external 3 PW q V V

externalW E K U 0

0

6 3externalW 3 10 4.2 10

The work done by the external force was negative, so the work done by the electric field was positive. The electric field “pulled” q3 in (keep in mind q2 is 4 times as big as q1).Positive work would have to be done by an external force to remove q3 from P.

J1026.1W 2external

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Example: find the total potential energy of the system of three charges.

q2q1

3 m

P

4 mx

y

q3

1 2 1 3 2 3

12 13 23

q q q q q qU = k + +

r r r

-6 -6 -6 -6 -6 -6

91×10 -4×10 1×10 3×10 -4×10 3×10

U = 9 10 + +4 3 5

J1026.1J109Wr

qqkJ1016.2U 23

external12

212 also

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Today’s agendum:





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The Electron Volt

An electron volt (eV) is the energy acquired by a particle of charge e when it moves through a potential difference of 1 volt.

U= qV

-191 eV= 1.6 10 C 1 V

-191 eV= 1.6 10 J

This is a very small amount of energy on a macroscopic scale, but electrons in atoms typically have a few eV (10’s to 1000’s) of energy.

32

Today’s agendum:

Electric potential of a charge distribution.You must be able to calculate the electric potential for a charge distribution.

Equipotentials.You must be able to sketch and interpret equipotential plots.

Potential gradient.You must be able to calculate the electric field if you are given the electric potential.

Potentials and fields near conductors.You must be able to use what you have learned about electric fields, Gauss’ “Law”, and electric potential to understand and apply several useful facts about conductors in electrostatic equilibrium.

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Electric Potential of a Charge Distribution

Example: potential and electric field between two parallel conducting plates.

Assume V0<V1 (so we can determine the direction of the electric field). Also assume the plates are large compared to their separation, so the electric field is constant and perpendicular to the plates.

V0 V1

EAlso, let the plates be separated by a distance d.

d

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V0 V1

plate 1

1 0 plate 0V V V E d

E

d

d d

0 0V E dx E dx Ed

d

xy

z

VE , or V Ed

d

I’ll discuss in lecture why the absolute value signs are needed.

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V Ed

Important note: the derivation of

did not require rectangular plates, or any plates at all. It works as long as E is uniform.

In general, E should be replaced by the component of along the displacement vector .

E

d

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Example: A rod of length L located along the x-axis has a total charge Q uniformly distributed along the rod. Find the electric potential at a point P along the y-axis a distance d from the origin.

y

x

P

d

L

dq

xdx

r

=Q/L

dq=dx

2 2

dq dxdV k k

r x d

L

0V dV

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y

x

P

d

L

dq

xdx

r

L L

2 2 2 20 0

dx Q dxV k k

Lx d x d

A good set of math tables will have the integral:

2 2

2 2

dxln x x d

x d

2 2kQ L L dV ln

L d

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Example: Find the electric potential due to a uniformly charged ring of radius R and total charge Q at a point P on the axis of the ring.

P

R

dq

r

xx

Every dq of charge on the ring is the same distance from the point P.

2 2

dq dqdV k k

r x R

2 2ring ring

dqV dV k

x R

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P

R

dq

r

xx

2 2 ring

kV dq

x R

2 2

kQV

x R

Could you use this expression for V to calculate E? Would you get the same result as we obtained previously?

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23

22ring,x

Rx

kxQE

P

a

dq

r

x

dE

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Example: A disc of radius R has a uniform charge per unit area and total charge Q. Calculate V at a point P along the central axis of the disc at a distance x from its center.

P

r

dq

xxR

The disc is made of concentric rings. The area of a ring at a radius r is 2rdr, and the charge on each ring is (2rdr).

We can use the equation for the potential due to a ring, replace R by r, and integrate from r=0 to r=R.

ring 2 2

k 2 rdrdV

x r

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P

r

dq

xxR

R

2 2 2 2ring ring 00 0

1 2 rdr rdrV dV

4 2x r x r

R

2 2 2 2 2 22

0 0 00

QV x r x R x x R x

2 2 2 R

2

Q

R

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2 22

0

QV x R x

2 R

P

r

dq

xxR

Could you use this expression for V to calculate E?

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See your text for other examples of potentials calculated from charge distributions, as well as an alternate discussion of the electric field between charged parallel plates.

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PRACTICAL APPLICATION

For some reason you think practical applications are important.Well, I found one!

http://www.electricmuseum.com/exhibits/cook/

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Today’s agendum:





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Equipotentials

Equipotentials are contour maps of the electric potential.

http://www.omnimap.com/catalog/digital/topo.htm

http://www.omnimap.com/catalog/digital/topo.htm

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The electric field must be perpendicular to equipotential lines. Why?

Otherwise work would be required to move a charge along an equipotential surface, and it would not be equipotential.

In the static case (charges not moving) the surface of a conductor is an equipotential surface. Why?

Otherwise charge would flow and it wouldn’t be a static case.

Equipotential lines are another visualization tool. They illustrate where the potential is constant. Equipotential lines are actually projections on a 2-dimensional page of a 3-dimensional equipotential surface. (“Just like” the contour map.)

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Here are some electric field and equipotential lines I generated using an electromagnetic field program.

Equipotential lines are shown in red.

50

Today’s agendum:





51

Potential Gradient(Determining Electric Field from Potential)

The electric field vector points from higher to lower potentials.More specifically, E points along shortest distance from a higher equipotential surface to a lower equipotential surface.You can use E to calculate V:

b

b a aV V E d .

You can use the differential version of this equation to calculate E from a known V:

dVdV E d E d E

d

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For spherically symmetric charge distribution:

r

dVE

dr

In one dimension:

x

dVE

dx

In three dimensions:

x y z

V V VE , E , E .

x y z

V V Vˆ ˆ ˆor E i j k Vx y z

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r

dVE

dr

x y z

V V VE , E , E .

x y z

dVE

d

Calculate -dV/d(whatever) including all signs. If the result is +, E vector points along the +(whatever) direction. If the result is -, E vector points along the –(whatever) direction.

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Example: In a region of space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx, where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are constants. Find the electric field at the origin.

2x

(0,0,0)(0,0,0)

VE (0,0,0) Ay 2Bx C C

x

y (0,0,0)(0,0,0)

VE (0,0,0) (2Axy) 0

y

z(0,0,0)

VE (0,0,0) 0

z

V ˆE(0,0,0) 400 im

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Today’s agendum:





56

When there is a net flow of charge inside a conductor, the physics is generally complex.

Potentials and Fields Near Conductors

When there is no net flow of charge, or no flow at all (the electrostatic case), then a number of conclusions can be reached using Gauss’ “Law” and the concepts of electric fields and potentials…

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The electric field inside a conductor is zero.

Summary of key points (electrostatic case):

Any net charge on the conductor lies on the outer surface.The potential on the surface of a conductor, and everywhere inside, is the same.

The electric field just outside a conductor must be perpendicular to the surface.

Equipotential surfaces just outside the conductor must be parallel to the conductor’s surface.

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Another key point: the charge density on a conductor surface will vary if the surface is irregular, and surface charge collects at “sharp points.”

Therefore the electric field is large (and can be huge) near “sharp points.”

Another Practical Application

To best shock somebody, don’t touch them with your hand; touch them with your fingertip.

Better yet, hold a small piece of bare wire in your hand and gently touch them with that.

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Today’s agendum:

Capacitance.You must be able to apply the equation C=Q/V.

Capacitors: parallel plate, cylindrical, spherical.You must be able to calculate the capacitance of capacitors having these geometries, and you must be able to use the equation C=Q/V to calculate parameters of capacitors.

Circuits containing capacitors in series and parallel.You must be understand the differences between, and be able to calculate the “equivalent capacitance” of, capacitors connected in series and parallel.

60

Capacitors and Dielectrics

Capacitance

A capacitor is basically two parallel conducting plates with air or insulating material in between.

V0 V1

E

L

A capacitor doesn’t have to look like metal plates.

Capacitor for use in high-performance audio systems.

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When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates.

Capacitor plates build up charge.

The battery in this circuit has some voltage V. We haven’t discussed what that means yet.

The symbol representing a capacitor in an electric circuit looks like parallel plates. Here’s the symbol for a battery, or an external potential. +-

-

-V

62

In a battery, chemical reactions release energy. The energy is expendedby transporting charged particles from one terminal of the batteryto the other. A “charge separation” is maintained.

63

If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy).

Capacitors are also very good at releasing their stored charge all at once. The capacitors in your tube-type TV are so good at storing energy that touching the two terminals at the same time can be fatal, even though the TV may not have been used for months.High-voltage TV capacitors are supposed to have “bleeder resistors” that drain the charge away after the circuit is turned off. I wouldn’t bet my life on it.

Graphic from http://www.feebleminds-gifs.com/.

+

+

-

-V

conducting wires

http://www.feebleminds-gifs.com/

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assortment of capacitors

65

The magnitude of charge acquired by each plate of a capacitor is Q=CV where C is the capacitance of the capacitor.

The unit of C is the farad but most capacitors have values of C ranging from picofarads to microfarads (pF to F).

micro 10-6, nano 10-9, pico 10-12 (Know for exam!)

QC

V C is always

positive.

+Q

+

-Q

-V

CHere’s this V again. It is the potential difference provided by the “external potential”. For example, the voltage of a battery.

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Today’s agendum:




67

Parallel Plate Capacitance

V0 V1

E

d

We previously calculated the electric field between two parallel charged plates:

0 0

QE .

A

This is valid when the separation is small compared with the plate dimensions. We also showed that E and V are related:

+Q-Q

A

d d

0 0V E d E dx Ed .

0

0

AQ Q QC

V Ed dQd

A

This lets us calculate C for a parallel plate capacitor.

68

Reminders:Q

CV

Q is the magnitude of the charge on either plate.

V is actually the magnitude of the potential difference between the plates. V is really |V|. Your book calls it Vab.

C is always positive.

69

V0 V1

E

d

+Q-Q

A

0ACd

Parallel plate capacitance depends “only” on geometry.

This expression is approximate, and must be modified if the plates are small, or separated by a medium other than a vacuum.

0ACd

Greek letter Kappa. For today’s lecture use Kappa=1.

70

Isolated Sphere Capacitance

An isolated sphere can be thought of as concentric spheres with the outer sphere at an infinite distance and zero potential.We already know the potential outside a conducting sphere:

0

QV .

4 r

The potential at the surface of a charged sphere of radius R is

0

QV

4 R

so the capacitance at the surface of an isolated sphere is

0

QC 4 R.

V

71

Capacitance of Concentric Spheres

Let’s calculate the capacitance of a concentric spherical capacitor of charge Q.

In between the spheres

20

QE

4 r

b

2a0 0

Q dr Q 1 1V

4 r 4 a b

04QC

1 1Va b

You need to do this derivation if you have a problem on spherical capacitors!

+Q

-Q

b

a

72

04QC

1 1Va b

Let aR and b to get the capacitance of an isolated sphere.

+Q

-Q

b

a

alternative calculation of capacitance of isolated sphere

73

We can also calculate the capacitance of a cylindrical capacitor (made of coaxial cylinders).

L

Coaxial Cylinder Capacitance

The next slide shows a cross-section view of the cylinders.

74

Q

-Q

br

a

E

d

Gaussian surface

Q λ L λ LC = = =

bΔV ΔV2k λ ln

a

02πε LLC = =

b b2k ln ln

a a

Lowercase c is capacitance per unit length: 02πεc =

bln

a

2kλE =

r

This derivation is sometimes needed for homework problems!

b b

b a r

a a

ΔV = V -V = - E d = - E dr

b

a

dr bΔV = - 2k λ = - 2k λ ln

r a

75

Example: calculate the capacitance of a capacitor whose plates are 20 cm x 3 cm and are separated by a 1.0 mm air gap.

d = 0.001area = 0.2 x 0.03

If you keep everything in SI (mks) units, the result is “automatically” in SI units.

0ACd

128.85 10 0.2 0.03C

0.001

12C 53 10 F

C 53 pF

76

Example: what is the charge on each plate if the capacitor is connected to a 12 volt* battery?

0 V

+12 V

V= 12V

Q CV

12Q 53 10 12

10Q 6.4 10 C

*Remember, it’s the potential difference that matters.


77

Example: what is the electric field between the plates?

0 V

+12 V

V= 12V

d = 0.001

E

VE

d

12VE

0.001 m

VE 12000 ,"up."

m


78

Today’s agendum:




79

Capacitors in Circuits

Recall: this is the symbol representing a capacitor in an electric circuit.And this is the symbol for a battery… +-

…or this…

…or this.

80

Capacitors connected in parallel:C1

C2

C3

+ -

V

The potential difference (voltage drop) from a to b must equal V.

a b

Vab = V = voltage drop across each individual capacitor.

Vab

Circuits Containing Capacitors in Parallel

Note how I have introduced the idea that when circuit components are connected in parallel, then the voltage drops across the components are all the same. You may use this fact in homework solutions.

81

C1

C2

C3

+ -

V

a

Q = C V

Q1 = C1 V

& Q2 = C2 V

& Q3 = C3 V

Now imagine replacing the parallel combination of capacitors by a single equivalent capacitor.

By “equivalent,” we mean “stores the same total charge if the voltage is the same.”

Ceq

+ -

V

a

Q1 + Q2 + Q3 = Ceq V = Q

Q3

Q2

Q1

+ -

Q

Important!

82

Q1 = C1 V Q2 = C2 V Q3 = C3 V

Q1 + Q2 + Q3 = Ceq V

Summarizing the equations on the last slide:

Using Q1 = C1V, etc., gives

C1V + C2V + C3V = Ceq V

C1 + C2 + C3 = Ceq (after dividing both sides by V)

Generalizing:

Ceq = Ci (capacitors in parallel)

C1

C2

C3

+ -

V

a b

83

Capacitors connected in series:

C1 C2

+ -

V

C3

An amount of charge +Q flows from the battery to the left plate of C1. (Of course, the charge doesn’t all flow at once).

+Q -Q

An amount of charge -Q flows from the battery to the right plate of C3. Note that +Q and –Q must be the same in magnitude but of opposite sign.

Circuits Containing Capacitors in Series

84

C1 C2

+ -

V

C3

+QA

-QB

The charges +Q and –Q attract equal and opposite charges to the other plates of their respective capacitors:

-Q +Q

These equal and opposite charges came from the originally neutral circuit regions A and B.

Because region A must be neutral, there must be a charge +Q on the left plate of C2.

Because region B must be neutral, there must be a charge --Q on the right plate of C2.

+Q -Q

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C1 C2

+ -

V

C3

+QA

-QB

-Q +Q+Q -Q

Q = C1 V1 Q = C2 V2 Q = C3 V3

The charges on C1, C2, and C3 are the same, and are

But we don’t know V1, V2, and V3 yet.

a b

We do know that Vab = V and also Vab = V1 + V2 + V3.

V3V2V1

Vab

Note how I have introduced the idea that when circuit components are connected in series, then the voltage drop across all the components is the sum of the voltage drops across the individual components. This is actually a consequence of the conservation of energy.

86

Ceq

+ -

V

+Q -QV

Let’s replace the three capacitors by a single equivalent capacitor.

By “equivalent” we mean V is the same as the total voltage drop across the three capacitors, and the amount of charge Q that flowed out of the battery is the same as when there were three capacitors.

Q = Ceq V

87

Collecting equations:

Q = C1 V1 Q = C2 V2 Q = C3 V3

Vab = V = V1 + V2 + V3.

Q = Ceq V

Substituting for V1, V2, and V3:1 2 3

Q Q QV = + +

C C C

Substituting for V:eq 1 2 3

Q Q Q Q = + +

C C C C

Dividing both sides by Q:eq 1 2 3

1 1 1 1 = + +

C C C C

Important!

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Generalizing:

(capacitors in series)ieq i

1 1 =

C C

89

C3

C2

C1

I don’t see a series combination of capacitors, but I do see a parallel combination.

C23 = C2 + C3 = C + C = 2C

Example: determine the capacitance of a single capacitor that will have the same effect as the combination shown. Use C1 = C2 = C3 = C.

90

C1= CC23 = 2C

Now I see a series combination.

eq 1 23

1 1 1 = +

C C C

eq

1 1 1 2 1 3 = + = + =

C C 2C 2C 2C 2C

eq

2C = C

3

91

Example: for the capacitor circuit shown, C1 = 3F, C2 = 6F, C3 = 2F, and C4 =4F. (a) Find the equivalent capacitance. (b) if V=12 V, find the potential difference across C4.

I’ll work this at the blackboard.

C3

C2C1 C4

V

92

Overview

Electric charge and electric forceCoulomb’s Law

Electric fieldcalculating electric fieldmotion of a charged particle in an electric field

Gauss’ “Law”electric fluxcalculating electric field using Gaussian surfacesproperties of conductors

93

Overview

Electric potential and electric potential energycalculating potentials and potential energycalculating fields from potentialsequipotentialspotentials and fields near conductors

Capacitorscapacitance of parallel plates, concentric

cylinders, (concentric spheres not for this exam)equivalent capacitance of capacitor network

Don’t forget the concepts from Physics 103 that were frequently used!

94

F1

F2

+Q

-Q+Q

P

a

x

y

1 2

1 2

ˆF F cos F cos i

ˆ F sin F sin j

2 2

2 2

2 2

2 2

kQ kQ ˆF cos 60 cos 60 ia a

kQ kQ ˆ sin 60 sin 60 ja a

Three charges +Q, +Q, and –Q, are located at the corners of an equilateral triangle with sides of length a. What is the force on the charge located at point P (see diagram)?

95

F1

F2

+Q

-Q+Q

P

a

x

y

2

2

kQ ˆF 2 cos 60 ia

2

2

kQ ˆF ia

96

F1

F2

+Q

-Q+Q

P

a

x

y

You can repeat the above calculation,replacing F by E.

Or…F

Eq

2

2

kQ ˆ2 cos 60 iaE

Q

2

kQ ˆE 2 cos60 ia

What is the electric field at P due to the two charges at the base of the triangle?

97

An insulating spherical shell has an inner radius a and outer radius b. The shell has a total charge Q and a uniform charge density. Find the magnitude of the electric field for r<a.

0

o

enclosedqAdE

98

3 3

2

o

4 4r a

3 3E 4 r

3 34 4Q b a

3 3

An insulating spherical shell has an inner radius a and outer radius b. The shell has a total charge Q and a uniform charge density. Find the magnitude of the electric field for a<r<b.

o

enclosedqAdE

99

f iV V V Ed

f i other i fE E E W .

U q V

An electron has a speed v. Calculate the magnitude and direction of an electric field that will stop this electron in a distance D.

Do you understand that these E’s have different meanings?

Use magnitudes if you have determined direction of E by other means.

Know where this comes from!

100

QQ

P

a

Two equal positive charges Q are located at the base of an equilateral triangle with sides of length a. What is the potential at point P (see diagram)?

101

QQ

P

a

Q

Three equal positive charges Q are located at the corners of an equilateral triangle with sides of length a. What is the potential energy of the charge located at point P (see diagram)?

102

For the capacitor system shown, C1=6.0 F, C2=2.0 F, and C3=10.0 F. (a) Find the equivalent capacitance.

V0

C1

C2 C3

103

For the capacitor system shown, C1=6.0 F, C2=2.0 F, and C3=10.0 F. The charge on capacitor C3 is found to be 30.0 C. Find V0.

V0

C1

C2 C3

104

Today’s agendum:

Energy Storage in Capacitors.You must be able to calculate the energy stored in a capacitor, and apply the energy storage equations to situations where capacitor configurations are altered.

Dielectrics.You must understand why dielectrics are used, and be able include dielectric constants in capacitor calculations.

105

Energy Storage in Capacitors

Let’s calculate how much work it takes to charge a capacitor.The work required for an external force to move a charge dq through a potential difference V is dW = dq V.

V+ -

+q -q

+dq

From Q=CV ( V = q/C):

qdW dq

C

q is the amount of charge on the capacitor during the time the charge dq is being moved.

We start with zero charge on the capacitor, and end up with Q, so

Q2 2Q Q

0 00

q q QW dW dq .

C 2C 2C

106

The work required to charge the capacitor is the amount of energy you get back when you discharge the capacitor (because the electric force is conservative).

Thus, the work required to charge the capacitor is equal to the potential energy stored in the capacitor.

2QU .

2C

Because C, Q, and V are related through Q=CV, there are three equivalent ways to write the potential energy.

2 2Q CV QVU .

2C 2 2

107

2 2Q CV QVU .

2C 2 2

All three equations are valid; use the one most convenient for the problem at hand.

It is no accident that we use the symbol U for the energy stored in a capacitor. It is just another “version” of electrical potential energy. You can use it in your energy conservation equations just like any other form of potential energy!

108

Example: a camera flash unit stores energy in a 150 F capacitor at 200 V. How much electric energy can be stored?

2CVU

2

6 2150 10 200U

2

U 3 J


109

Example: compare the amount of energy stored in a capacitor with the amount of energy stored in a battery.A 12 V car battery rated at 100 ampere-hours stores 3.6x105 C of charge and can deliver at least 4.3x106 joules of energy (we’ll learn how to calculate that later in the course).

If a battery stores so much more energy, why use capacitors?

106 joules of energy are stored at high voltage in capacitor banks, and released during a period of a few milliseconds. The enormous current produces incredibly high magnetic fields.

A 100 F capacitor that stores 3.6x105 C at 12 V stores an amount of energy U=CV2/2=7.2x10-3 joules.If you want your

capacitor to store lots of energy, store it at a high voltage.

http://www.lanl.gov/mst/nhmfl/

110

Energy Stored in Electric Fields

V+ -

+Q -Q

E

21U C V

2

d

area A

20A1U Ed

2 d

20

1U Ad E

2

Energy is stored in the capacitor:

The “volume of the capacitor” is Volume=Ad

111

Energy stored per unit volume (u):

V+ -

+Q -Q

E

d

area A

20

20

1Ad E 12u EAd 2

The energy is “stored” in the electricfield!

We’ve gone from the concrete (electric charges experience forces)……to the abstract (electric charges create electric fields)…

…to an application of the abstraction (electric field contains energy).

112

20

1u E

2 V

+ -

+Q -Q

E

f

area A

This is not a new “kind” of energy. It’s the electric potential energy resulting from the coulomb force between charged particles.

Or you can think of it as the electric energy due to the field created by the charges. Same thing.

“The energy in electromagnetic phenomena is the same as mechanical energy. The only question is, ‘Where does it reside?’ In the old theories, it resides in electrified bodies. In our theory, it resides in the electromagnetic field, in the space surrounding the electrified bodies.”—James Maxwell

113

Today’s agendum:

Energy Storage in Capacitors.You must be able to calculate the energy stored in a capacitor, and apply the energy storage equations to situations where capacitor configurations are altered.

Dielectrics.You must understand why dielectrics are used, and be able include dielectric constants in capacitor calculations.

114

If an insulating sheet (“dielectric”) is placed between the plates of a capacitor, the capacitance increases by a factor , which depends on the material in the sheet. is the dielectric constant of the material.

dielectric

In general, C = 0A / d. is 1 for a vacuum, and 1 for air. (You can also define = 0 and write C = A / d).

Dielectrics

AC = .

d

115

The dielectric is the thin insulating sheet in between the plates of a capacitor.

dielectric

Any reasons to use a dielectric in a capacitor?

Lets you apply higher voltages (so more charge).

Lets you place the plates closer together (make d smaller).

Increases the value of C because >1.

Q=CV

AC =

d

Makes your life as a physics student more complicated.

116

Example: a parallel plate capacitor has an area of 10 cm2 and plate separation 5 mm. 300 V is applied between its plates. If neoprene is inserted between its plates, how much charge does the capacitor hold.

V=300 Vd=5 mm

A=10 cm2

=6.7

-12 -4

-3

6.7 8.85×10 10×10C =

5×10

Q=CV

-11C=1.19 10 F

-11 -9Q= 1.19 10 300 3.56 10 C =3.56 nC

AC =

d

117

Example: how much charge would the capacitor on the previous slide hold if the dielectric were air?

The calculation is the same, except replace 6.7 by 1.

Or just divide the charge on the previous page by 6.7 to get.Q=0.53 nC V=300

Vd=5 mm

A=10 cm2

=1

118

A capacitor connected as shown acquires a charge Q. V

While the capacitor is still connected to the battery, a dielectric material is inserted.

Will Q increase, decrease, or stay the same?Why?

V

V=0Conceptual Example

119

Example: find the energy stored in the capacitor in slide 13.

V=300 Vd=5 mm

A=10 cm2

=6.7

-11C=1.19 10 F

1

2 2

U= C V

1

2 2-11U= 1.19 10 300

-7U=5.36 10 J

120

Example: the battery is now disconnected. What are the charge, capacitance, and energy stored in the capacitor?

V=300 Vd=5 mm

A=10 cm2

=6.7

The charge and capacitance are unchanged, so the voltage drop and energy stored are unchanged.Q =3.56 nC

-11C=1.19 10 F

-7U=5.36 10 J

121

Example: the dielectric is removed without changing the plate separation. What are the capacitance, charge, potential difference, and energy stored in the capacitor?

V=300 Vd=5 mm

A=10 cm2

=6.7 -12 -4

-3

8.85×10 10×10C =

5×10

-12C=1.78 10 F

AC =

d

V=?d=5 mm

122

V=?d=5 mm

A=10 cm2

Q =3.56 nC


The charge remains unchanged, because there is nowhere for it to go.

123

V=?d=5 mm

A=10 cm2

-9

-12

3.56 10QV = =

C 1.78 10

V = 2020 V


Knowing C and Q we can calculate the new potential difference.

124

V=2020 Vd=5 mm

A=10 cm2 1

2 2

U= C V

1

2 2-12U= 1.78 10 2020

-6U=3.63 10 J


125

-7beforeU =5.36 10 J

-6afterU =3.63 10 J

after

before

U=6.7

U

Huh?? The energy stored increases by a factor of ??

Sure. It took work to remove the dielectric. The stored energy increased by the amount of work done.

externalU=W

1. 2 Today’s agendum: Electric potential energy. You must be able to use electric potential energy...

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