Week 5

Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is

1. zero

2. between zero and 90°

3. 90°

4. not enough information given to decide

Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is

1. zero

2. between zero and 90°

3. 90°

4. not enough information given to decide

Since there is no change in V along the equipotential surface

ΔV = −E⋅dr = 0But this means that the vectors E and dr are perpendicular to each other.

What is the electric potential created by these three batteriesconnected as shown?

1. 1 Volt

2. 2 Volt

3. 3 Volt

4. 4 Volt

5. 5 Volt

6. 6 Volt

7. 7 Volt

8. 8 Volt

What is the electric potential created by these three batteriesconnected as shown?

1. 1 Volt

2. 2 Volt

3. 3 Volt

4. 4 Volt

5. 5 Volt

6. 6 Volt

7. 7 Volt

8. 8 Volt

Which potential-energy graph describes this electric field?

1 2 3 4 5

Which potential-energy graph describes this electric field?

1 2 3 4 5

The direction of the electric potential gradient at a certain point .....

1. is the same as the direction of the electric field at that point

2. is opposite to the direction of the electric field at that point

3. is perpendicular to the direction of the electric field at that point

4. not enough information given to decide

The direction of the electric potential gradient at a certain point .....

1. is the same as the direction of the electric field at that point

2. is opposite to the direction of the electric field at that point

3. is perpendicular to the direction of the electric field at that point

4. not enough information given to decide

!E = −!∇V

Understand first a simpler analog:topographic maps have “contour lines” that are thelocus of points such that ground elevation is constant:

h(!r) = constant

When you follow a path on a topographic map that crosses contour lines, you are either climbing or descending. A path running parallel to contour lines is flat.

DENSE contour lines mean STEEP terrain and SPARSE ones mean FLAT terrain.

The “gradient” is in the direction of steepest ascent and its size is the rate of ascent.

Which set of equipotential surfaces matches this electric field?

1. 2. 3.

4. 5.

Which set of equipotential surfaces matches this electric field?

1. 2. 3.

4. 5.

Example: Determine the electric field of a dipole, far away from the dipole

V = kpcos θ

r2= kp

x

r3Use the result from last week:

(COB)

Ans:

Ex = −kpr2 − 3x2

r5= −kp

1− 3 cos2 θ

r3

Ey = kp3xy

r5= 3kp

sin θ cos θ

r3

1. V1 = V2 = V3 and E1 = E2 = E3

2. V1 = V2 = V3 and E1 > E2 > E3

3. V1 > V2 > V3 and E1 = E2 = E3

4. V1 > V2 > V3 and E1 > E2 > E3

5. V3 > V2 > V1 and E1 = E2 = E3

Three charged, metal spheres of

different radii are connected by

a thin metal wire. The potential

and electric field at the surface

of each sphere are V and E.

Which of the following is true?

1. V1 = V2 = V3 and E1 = E2 = E3

2. V1 = V2 = V3 and E1 > E2 > E3

3. V1 > V2 > V3 and E1 = E2 = E3

4. V1 > V2 > V3 and E1 > E2 > E3

5. V3 > V2 > V1 and E1 = E2 = E3

Three charged, metal spheres of

different radii are connected by

a thin metal wire. The potential

and electric field at the surface

of each sphere are V and E.

Which of the following is true?

Capacitors

A system consisting of two conductors (“plates”)

Now, equal magnitude, opposite sign charge, ±Q , produces field between them

Hence, potential difference V

The more charge Q , the larger V

This is a linear relation (V ∝ Q)

(slide here really to remind me of demo 1st)

(a)

(b)

+ +

+++

+– –

––––

V =1C

Q

The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to –2Q, while keeping the conductors in the same positions.

What effect does this have on the capacitance C?

1. C is multiplied by a factor of 4

2. C is multiplied by a factor of 2

3. C is unchanged

4. C is multiplied by a factor of 1/2

5. C is multiplied by a factor of 1/4

The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to –2Q, while keeping the conductors in the same positions.

What effect does this have on the capacitance C?

1. C is multiplied by a factor of 4

2. C is multiplied by a factor of 2

3. C is unchanged

4. C is multiplied by a factor of 1/2

5. C is multiplied by a factor of 1/4

You reposition the two plates of a capacitor so that the capacitance doubles.

If the charges +Q and –Q on the two plates are kept constant in this process, what happens to the potential difference Vab between the two plates?

1. Vab is multiplied by a factor of 4

2. Vab is multiplied by a factor of 2

3. Vab is unchanged

4. Vab is multiplied by a factor of 1/2

5. Vab is multiplied by a factor of 1/4

You reposition the two plates of a capacitor so that the capacitance doubles.

If the charges +Q and –Q on the two plates are kept constant in this process, what happens to the potential difference Vab between the two plates?

1. Vab is multiplied by a factor of 4

2. Vab is multiplied by a factor of 2

3. Vab is unchanged

4. Vab is multiplied by a factor of 1/2

5. Vab is multiplied by a factor of 1/4

Example: our demod is approx 0.5mm, and A is approx 1.0 m2

C ≈ 1.8× 10−8 F = 18 nF

so how much charge did we

put in? Q = CV ≈ 1.8× 10−5 Cfor 1 kV:

(COB)

(This ignores the dielectric constant of the mica)

Capacitor C1 is connected

across a battery of 5 V. An

identical capacitor C2 is

connected across a battery of 10

V. Which one has the most

charge?

1) C1

2) C2

3) both have the same charge

4) it depends on other factors

+Q –Q

Since Q = C V and the two capacitors are

identical, the one that is connected to the

greater voltage has the most charge,

which is C2 in this case.

Capacitor C1 is connected

across a battery of 5 V. An

identical capacitor C2 is

connected across a battery of 10

V. Which one has the most

charge?

1) C1

2) C2

3) both have the same charge

4) it depends on other factors

+Q –Q

1) increase the area of the plates

2) decrease separation between the plates

3) decrease the area of the plates

4) either (1) or (2)

5) either (2) or (3)

What must be done to

a capacitor in order to

increase the amount of

charge it can hold (for

a constant voltage)?

+Q –Q

Since Q = C V, in order to increase the charge

that a capacitor can hold at constant voltage,

one has to increase its capacitance. Since the

capacitance is given by , that can be

done by either increasing A or decreasing d.

1) increase the area of the plates

2) decrease separation between the plates

3) decrease the area of the plates

4) either (1) or (2)

5) either (2) or (3)

What must be done to

a capacitor in order to

increase the amount of

charge it can hold (for

a constant voltage)?

+Q –Q

+Q –Q

A parallel-plate capacitor

initially has a voltage of 400 V

and stays connected to the

battery. If the plate spacing is

now doubled, what happens?

1) the voltage decreases

2) the voltage increases

3) the charge decreases

4) the charge increases

5) both voltage and charge change

Since the battery stays connected, the

voltage must remain constant ! Since

when the spacing d is

doubled, the capacitance C is halved.

And since Q = C V, that means the

charge must decrease.

+Q –Q

A parallel-plate capacitor

initially has a voltage of 400 V

and stays connected to the

battery. If the plate spacing is

now doubled, what happens?

1) the voltage decreases

2) the voltage increases

3) the charge decreases

4) the charge increases

5) both voltage and charge change

Follow-up: How do you increase the charge?

A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?

1) 100 V

2) 200 V

3) 400 V

4) 800 V

5) 1600 V

+Q –Q

Once the battery is disconnected, Q has to

remain constant, since no charge can flow

either to or from the battery. Since

when the spacing d is doubled, the

capacitance C is halved. And since Q = C V,

that means the voltage must double.

A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?

1) 100 V

2) 200 V

3) 400 V

4) 800 V

5) 1600 V

+Q –Q

The textbook calculates the capacitance of concentric spheres. Do another example:

co-axial cylindrical conductors

b

a

L

V (a)− V (b) =λ

2πε0ln

(b

a

)

C =2πε0L

ln(b/a)

Use previous chapter result (also from homework):

(COB)

Again, notice:- only depends on geometry, - increases with size of plates, - increases as plates get close ( a → b )

Connecting circuit components in Parallel and in Series

Some circuit component

Two circuit elements connected in series:the charge of one necessarily flows into the other:

Two circuit elements connected in parallel:the potential differences across them are necessarily equal:

Equivalent capacitanceof capacitors in parallel

(a) (b)

C1 C2

C1

C2

(a) (b)

C1 C2

C1

C2

=

(COB)

C = C1 + C2

Equivalent capacitanceof capacitors in series

(a) (b)

C1 C2

C1

C2

=

(COB)(a) (b)

C1 C2

C1

C2

1C

=1C1

+1C2

+Q

–Q –

++++++

– – – – –

+

–

+

–E

+Q

–Q –

++++++

– – – – –

+

–

+

–E

Two capacitors are connected together as shown.

What is the equivalent capacitance of the two capacitors as a unit?

1. Ceq = 18 µF

2. Ceq = 9 µF

3. Ceq = 6 µF

4. Ceq = 4 µF

5. Ceq = 2 µF

Two capacitors are connected together as shown.

What is the equivalent capacitance of the two capacitors as a unit?

1. Ceq = 18 µF

2. Ceq = 9 µF

3. Ceq = 6 µF

4. Ceq = 4 µF

5. Ceq = 2 µF

Two capacitors are connected together as shown.

If the charge on the 12–µF capacitor is 24 microcoulombs (24 µC), what is the charge on the 6–µF capacitor?

1. 48 µC

2. 36 µC

3. 24 µC

4. 12 µC

5. 6 µC

Two capacitors are connected together as shown.

If the charge on the 12–µF capacitor is 24 microcoulombs (24 µC), what is the charge on the 6–µF capacitor?

1. 48 µC

2. 36 µC

3. 24 µC

4. 12 µC

5. 6 µC

C1

C2

C33.0 Fµ

C4

12.0 Fµ

1.0 Fµ

4.0 Fµ

A

B

(a)

C1

C2

C34

12.0 Fµ

4.0 Fµ

4.0 Fµ

A

B

(b)

C1 C2343.0 Fµ4.0 Fµ

A

B

(c)

7.0 Fµ

A

B

(d)

Example: Find the equivalent capacity of the circuit shown in Fig. (a)

If we apply a voltage of 100Vbetween points A and Bwhat is the voltage acrosscapacitor C2?

Ans: 25V

Rank in order, from largest to smallest, the equivalent capacitance (Ceq)a to (Ceq)d

of circuits a to d.

1. (Ceq)a > (Ceq)b = (Ceq)c > (Ceq)d

2. (Ceq)b > (Ceq)a = (Ceq)d > (Ceq)c

3. (Ceq)c > (Ceq)a = (Ceq)d > (Ceq)b

4. (Ceq)d > (Ceq)b = (Ceq)c > (Ceq)a

5. (Ceq)d > (Ceq)b > (Ceq)a > (Ceq)c

Rank in order, from largest to smallest, the equivalent capacitance (Ceq)a to (Ceq)d

of circuits a to d.

1. (Ceq)a > (Ceq)b = (Ceq)c > (Ceq)d

2. (Ceq)b > (Ceq)a = (Ceq)d > (Ceq)c

3. (Ceq)c > (Ceq)a = (Ceq)d > (Ceq)b

4. (Ceq)d > (Ceq)b = (Ceq)c > (Ceq)a

5. (Ceq)d > (Ceq)b > (Ceq)a > (Ceq)c

C=5 C=3+3=6 C=1/(1/3+1/3)=3/2 C=3+1/(1/4+1/4)=5

o

o

C CC

Ceq

1) Ceq = 3/2 C

2) Ceq = 2/3 C

3) Ceq = 3 C

4) Ceq = 1/3 C

5) Ceq = 1/2 C

What is the equivalent capacitance,

Ceq , of the combination below?

The 2 equal capacitors in series add

up as inverses, giving 1/2 C. These

are parallel to the first one, which

add up directly. Thus, the total

equivalent capacitance is 3/2 C.

o

o

C CC

Ceq

1) Ceq = 3/2 C

2) Ceq = 2/3 C

3) Ceq = 3 C

4) Ceq = 1/3 C

5) Ceq = 1/2 C

What is the equivalent capacitance,

Ceq , of the combination below?

1) V1 = V2

2) V1 > V2

3) V1 < V2

4) all voltages are zero

C1 = 1.0 µF C3 = 1.0 µF

C2 = 1.0 µF

10 V

How does the voltage V1 across

the first capacitor (C1) compare

to the voltage V2 across the

second capacitor (C2)?

1) V1 = V2

2) V1 > V2

3) V1 < V2

4) all voltages are zero

C1 = 1.0 µF C3 = 1.0 µF

C2 = 1.0 µF

10 V

The voltage across C1 is 10 V. The combined capacitors C2

+C3 are parallel to C1. The voltage across C2+C3 is also 10 V. Since C2 and C3 are in series, their voltages add. Thus the voltage across C2 and C3 each has to be 5 V,

which is less than V1.

How does the voltage V1 across

the first capacitor (C1) compare

to the voltage V2 across the

second capacitor (C2)?

C1 = 1.0 µF C3 = 1.0 µF

C2 = 1.0 µF

10 V

1) Q1 = Q2

2) Q1 > Q2

3) Q1 < Q2

4) all charges are zero

How does the charge Q1 on the first

capacitor (C1) compare to the

charge Q2 on the second capacitor

(C2)?

C1 = 1.0 µF C3 = 1.0 µF

C2 = 1.0 µF

10 V

We already know that the

voltage across C1 is 10 V

and the voltage across C2

and C3 each is 5 V. Since Q

= CV and C is the same for

all the capacitors, then since

V1 > V2 therefore Q1 > Q2.

1) Q1 = Q2

2) Q1 > Q2

3) Q1 < Q2

4) all charges are zero

How does the charge Q1 on the first

capacitor (C1) compare to the

charge Q2 on the second capacitor

(C2)?

Initially the switch is in position A and capacitors C2 and C3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor?

Ans: Q1 = 0.83 mC, Q2 = Q3 = 0.63 mC V1 = 55 V, V2 = 34 V, V3 = 21 V

You reposition the two plates of a capacitor so that the capacitance doubles.

If the charges +Q and –Q on the two plates are kept constant in this process, what happens to the energy stored in the capacitor?

1. increases by a factor of 4

2. increases by a factor of 2

3. the stored energy is unchanged

4. decreases by a factor of 2

5. decreases by a factor of 4

You reposition the two plates of a capacitor so that the capacitance doubles.

If the charges +Q and –Q on the two plates are kept constant in this process, what happens to the energy stored in the capacitor?

1. increases by a factor of 4

2. increases by a factor of 2

3. the stored energy is unchanged

4. decreases by a factor of 2

5. decreases by a factor of 4

You slide a slab of dielectric between the plates of a capacitor. As you do this, the charges on the plates remain constant.

1. the stored energy increases

2. the stored energy is unchanged

3. the stored energy decreases

4. not enough information given to decide

+Q –Q +Q –Q

What effect does this have on the energy stored in the capacitor?

You slide a slab of dielectric between the plates of a capacitor. As you do this, the charges on the plates remain constant.

1. the stored energy increases

2. the stored energy is unchanged

3. the stored energy decreases

4. not enough information given to decide

+Q –Q +Q –Q

What effect does this have on the energy stored in the capacitor?

Two 5.0-cm-diameter metal disks separated by a 0.50-mm-thick piece of Pyrex glass are charged to a potential difference of 1000 V. What are (a) the surface charge density on the disks, and(b) the surface charge density on the glass?

Ans: (a) 8.3 x 10-5 C/m2 (b) 1.4 x 10-5 C/m2

Look up in table 30.1, for Pyrex glass κ = 4.7