Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt,...

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip

Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar

Week 4. Potential, capacitance and capacitors – E from V • equipotential surfaces • E in and on a conductor • capacitors and capacitance • capacitors in series and in parallelSource: Halliday, Resnick and Krane, 5th Edition, Chaps. 28, 30.

Physics 2B for Materials and Structural EngineeringPhysics 2B for Materials and Structural Engineering

http://www.bgu.ac.il/atomchip

With 100,000 V on her body, why is this girl smiling???

E from V

Let U(r2) be the potential energy of a charge q at the point r2. We found that it is minus the work done by the electric force Fq in bringing the charge to r2:

If we divide both sides by q, we get (on the left side) potential instead of potential energy, and (on the right side) the electric field instead of the electric force:

. )()()(2

1

12 r

r

rrFrr dUU q

. )()()(2

1

12 r

r

rrErr dVV

E from V

Let V(r2) be the electric potential at the point r2. We have just obtained it from the electric field E:

Now let’s see how to obtain E from V. For r2 close to r1 we

can write r2 = r1 + Δr and expand V(r2) in a Taylor series:

. )()()(2

1

12 r

r

rrErr dVV

; ...)(

)()(

111

1

12

zz

Vy

y

Vx

x

VV

VV

zyxr

rrr

E from V

We can write this expansion more compactly using the “del” (gradient) operator:

so

with all the derivatives evaluated at the point r1. We also write

, ,,

zyx

, ,,)( 1

z

V

y

V

x

VV r

. z)( 1

z

Vy

y

Vx

x

VV rr

E from V

So now we can write

and therefore

...)()(

...z)()(

11

11

rrr

rrr

VV

z

Vy

y

Vx

x

VVV

. )()(

)()()(

)()()(

11

111

12

2

1

rrErr

rrErrr

rrErr

r

r

V

VV

dVV

E from V

The equation

is a scalar equation, but since we can vary each component of Δr independently, it actually yields three equations:

)()( 11 rrErr V

, )()(

, )()(

, )()(

11

11

11

rr

rr

rr

zz

yy

xx

EV

EV

EV

E from V

The equation

is a scalar equation, but since we can vary each component of Δr independently, it actually yields three equations, i.e.

)()( 11 rrErr V

, )()(

, )()(

, )()(

11

11

11

rr

rr

rr

z

y

x

EVz

EVy

EVx

E from V

which reduce to a single vector equation:

, )()(

, )()(

, )()(

11

11

11

rr

rr

rr

z

y

x

EVz

EVy

EVx

. )()( rrE V

E from V

Example 1 (Halliday, Resnick and Krane, 5th Edition, Chap. 28, Exercise 34): Rutherford discovered, 99 years ago, that an atom has a positive nucleus with a radius about 105 times smaller than the radius R of the atom. He modeled the electric potential inside the atom (r < R) as follows:

where Z is the atomic number (number of protons). What is the corresponding electric field?

, 22

31

4)(

2

2

0

R

r

Rr

ZerV

E from V

Answer:

since

, ˆ1

4

1

4

22

31

4)()(

220

220

2

2

0

r

rE

R

r

r

Ze

rR

r

r

Ze

R

r

Rr

ZerV

. ˆˆˆˆ

ˆˆˆ rrzyx

zyx

rr

zyx

z

r

y

r

x

rr

E from V

Example 2: Electric field of a dipole

We found that the electric potential V(x,y,z) of a dipole made of charges q at (0,0,d/2) and –q at (0,0,–d/2) is

z

–d/2

d/2

(x,y,z)

. )2

( re whe

, 1

1

4

),,(

222

0

dzyxr

rr

qzyxV

r+

r–

E from V

. )2

( re whe

, 2

2

8

)(

, 1

1

4

)(

, 1

1

4

)(

222

330

330

330

dzyxr

r

dz

r

dzq

z

VE

rr

qy

y

VE

rr

qx

x

VE

z

y

x

r

r

r

z

–d/2

d/2

(x,y,z)

r+

r–

E from V

Now we will consider the case d << r– , r+ and use these rules for 0 < α << 1 (derived from Taylor or binomial expansions) to approximate E:

. ...1) (1

, ... 1 1

1

, ... 2

1 1

nn

E from V

For d << r– , r+ we have

z

(x,y,z)

r+

r–

. where, 4

1

4

2

2221/2

2

2

2

2222

222

zyxrr

d

r

zdr

dzdzyx

dzyxr

d/2

–d/2

E from V

For d << r– , r+ we have

z

(x,y,z)

r+

r–

. 8

3

2

31

1

2

31

11

1

)(

1

so , 4

where, 1

2

2

2333/2

33

2

2

21/2

r

d

r

zd

rrrr

r

d

r

zdrr

d/2

–d/2

E from V

So

z

(x,y,z)

r–

now and ; 3

8

3

2

31

1

8

3

2

31

11

1

5

2

2

232

2

2333

r

zd

r

d

r

zd

rr

d

r

zd

rrr

d/2

–d/2

. 4

)3( ,

4

3 ,

4

35

0

22

50

50 r

qdrzE

r

yzqdE

r

xzqdE zyx

r+

E from V

We can check that these results coincide with the results we obtained for the special cases x = 0 = y and z = 0:

z

(x,y,z)d/2

–d/2 r+

. 4

, 2 3

03

0 r

qdE

z

qdE zz

. 4

)3( ,

4

3 ,

4

35

0

22

50

50 r

qdrzE

r

yzqdE

r

xzqdE zyx

r–

E from V

This may look complicated but it is easier than calculating Ex,

Ey and Ez directly!

z

(x,y,z)

r–d/2

–d/2 r+

. 4

)3( ,

4

3 ,

4

35

0

22

50

50 r

qdrzE

r

yzqdE

r

xzqdE zyx

Equipotential surfaces

We have already seen equipotential surfaces in pictures of lines of force:

and in connection with potential energy:r2

r1

r2

r1

rF(r)


All points on an equipotential surface are at the same electric potential.

Electric field lines and equipotential surfaces meet at right angles. Why? r2

r1

r2

r1

rF(r)


The surface of a conductor at electrostatic equilibrium – when all the charges in the conductor are at rest – is an equipotential surface, even if the conductor is charged:

Small pieces of thread (in oil) align with the electric field due to two conductors, one pointed and one flat, carrying opposite charges.

[From Halliday, Resnick and Krane]

y

V

x


We can also visualize the topography of the electric potential from the side (left) and from above (right) with equipotentials

as horizontal curves.


Quick quiz: In three space dimensions, rank the potential differences V(A) – V(B), V(B) – V(C), V(C) – V(D) and V(D) – V(E).

9 V

7 V

8 V

6 V

A

B

CDE

E in and on a conductor

Four rules about conductors at electrostatic equilibrium:

1. The electric field is zero everywhere inside a conductor.

Explanation: A conductor contains free charges (electrons) that move in response to an electric field; at electrostatic equilibrium, then, the electric field inside a conductor must vanish.

Example: An infinite conducting sheet ina constant electric field develops surface charges that cancel the electric field inside the conductor.

+

+

+

+

+

+

+

-

-

-

-

-

-

-




2. Any net charge on an isolated conductor lies on its surface.





Explanation: A charge inside the conductor would imply, via Gauss’s law, that the electric field could not be zero everywhere on a small surface enclosing it, contradicting 1.





3. The electric field on the surface of a conductor must be normal to the surface and equal to σ/ε0, where σ is the surface charge density.






Explanation: A component of E parallel to the surface would move free charges around. (Hence the surface of a conductor at electrostatic equilibrium is an equipotential surface, even if the conductor is charged.)


Let’s compare the electric fields due to two identical surface charge densities σ, one on a conductor with all the excess charge on one side (e.g. the outside of a charged sphere) and the other on a thin insulating sheet.

The figure shows a short “Gaussian can” straddling the thin charged sheet. If the can is short, we need to consider only the electric flux through the top and bottom.Gauss’s law gives 2EA = ФE = σA/ε0, so

E = σ/2ε0. But if there is flux only through

the top of the can, Gauss’s law gives EA = ФE = σA/ε0 and E = σ/ε0.

“Gaussian can”






4. On an irregularly shaped conductor, the surface charge density is largest where the curvature of the surface is largest.


Explanation: If we integrate E·dr along any electric field line, starting from the conductor and ending at infinity, we must get the same result, because the conductor and infinity are both equipotentials. But E drops more quickly from a sharp point or edge than from a smooth surface – as we have seen, E drops as 1/r2 from a point charge, as 1/r near a charged line, and scarcely drops near a charged surface. The integral far from the conductor is similar for all electric field lines; near the conductor, if E·dr drops more quickly from a sharp point or edge, it must be that E starts out larger there. Then, since E is proportional to the surface charge σ, it must be that σ, too, is larger at a sharp point or edge of a conductor.

Capacitors and capacitance

A capacitor is any pair of isolated conductors. We call the capacitor charged when one conductor has total charge q and the other has total charge –q. (But the capacitor is then actually neutral.)

Whatever the two conductors look like, the symbol for a capacitor is two parallel lines.

capacitor

battery

switch

http://tp01.com/etutorials/product_info.php?products_id=42&osCsid=0bbbf1c2b5e83b0420c4bd5c9836e4f1


Here is a circuit diagram with a battery to charge a capacitor, and a switch to open and close the circuit.

capacitor

battery

switch


The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV.

capacitor

battery

switch


The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV. (Note that q and ΔV both scale the same way as the electric field.)


The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV. (Note that q and ΔV both scale the same way as the electric field.)

The constant C is called the capacitance of the capacitor. The unit of capacitance is the farad F, which equals Coulombs per volt: F = C/V.


Example 1: Parallel-plate capacitor

If we can neglect fringing effects (that is, if we can take the area A of the conducting plates to be much larger than the distance d between the plates) then E = σ/ε0 where σ = q/A and

ΔV = Ed = qd/ε0A. By definition, q = C ΔV, hence C = ε0A/d

is the capacitance of an ideal parallel-plate capacitor.


Example 2: Cylindrical capacitor

Again, if we can neglect fringing effects (that is, if we can take the length L of the capacitor to be much larger than the inside radius b of the outer tube) then

. ln ln

2C

, )(ln 22

, , 2

1)(

0

00

0

ab

L

b/aL

q

r

dr

L

qV

b r a Lr

qrE

b

a

b

a


Example 3: Spherical capacitor

The diagram is unchanged, only E(r) is different:

.

4C

, 11

44

, , 4

1)(

0

02

0

20

ab

ab

ba

q

r

drqV

b r a r

qrE

b

a

b

a

Capacitors in series and in parallel

Combinations of capacitors have well-defined capacitances. Here are two capacitors in series:

capacitor

battery

switch



Since the potential across both capacitors is ΔV, we must have q1 = C1ΔV1 and q2 = C2ΔV2 where

ΔV1 + ΔV2 = ΔV. But the charge

on one capacitor comes from the other, hence q1 = q2 = q.



Since the potential across both capacitors is ΔV, we must have q = C1ΔV1 and q = C2ΔV2 where

ΔV1 + ΔV2 = ΔV. If the effective

capacitance is Ceff, then we have

hence

, 21

21eff C

q

C

qVVV

C

q

. 111

21eff CCC


For n capacitors in series, the generalized rule is

. 1

... 111

21eff nCCCC

C1 C2 Cn


Combinations of capacitors have well-defined capacitances. Here are two capacitors in parallel:

capacitor

battery

switch


Combinations of capacitors have well-defined capacitances. Here are two capacitors in parallel:

Since the potential across each capacitor is still ΔV, the charge on the capacitors is q1 = C1ΔV and q2 = C2ΔV. If the effective

capacitance is Ceff, then we have

C1ΔV + C2ΔV = q1 + q2 = CeffΔV,

thus capacitances in parallel add:

C1 + C2 = Ceff .


For n capacitors in parallel, the generalized rule is

Ceff = C1 + C2 +…+ Cn .C1

C2

Cn

Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:

The capacitors have identical capacitance C. What is the equivalent capacitance Ceff of each of these combinations?

A B

C D

A B

C D

Ceff = 3C

Ceff = C/3

Ceff = 2C/3

Ceff = 3C/2

Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:

The capacitors have identical capacitance C. What is the equivalent capacitance Ceff of each of these combinations?

Halliday, Resnick and Krane, 5th Edition, Chap. 30, Prob. 9:

Find the charge on each capacitor (a) with the switch open and(b) with the switch closed.

ΔV =12 V

C1 = 1 μF C2 = 2 μF

C3 = 3 μF C4 = 4 μF

Answer (a): We have C1ΔV1 = q1 and C3ΔV3 = q1 since those

two capacitors are equally charged. Now ΔV1 + ΔV3 = 12 V

and so q1/C1 + q1/C3 = 12 V and we solve to get q1 = q3 = 9 μC.

In just the same way we obtain q2 = q4 = 16 μC.

ΔV =12 V

C1 = 1 μF C2 = 2 μF

C3 = 3 μF C4 = 4 μF

Answer (b): We have C1ΔV12 = q1 and C2ΔV12 = q2 and in the

same way C3ΔV34 = q3 and C4ΔV34 = q4. Two more equations:

ΔV12 + ΔV34 = 12 V and q1 + q2 = q3 + q4. We solve these six

equations to obtain ΔV12 = 8.4 V and ΔV34 = 3.6 V, and charges

q1 = 8.4 μC, q2 = 16.8 μC, q3 =10.8 μC and q4 = 14.4 μC.

ΔV =12 V

C1 = 1 μF C2 = 2 μF

C3 = 3 μF C4 = 4 μF

Date post:	20-Dec-2015
Category:	Documents
View:	215 times
Download:	0 times

Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt,...

Documents