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Electric Potential (III) Text sections 25.5, 25.6

Fields, potential, and conductors

Practice: Chapter 25, problems 29, 33, 35, 39, 57, 61 Read page 710, Van de Graaff generator

We can use two completely different methods:

1. 

2.  Find from Gauss’s Law, then…

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L + + + + + + + + + + + + + + + + + O b

Total charge Q, uniform linear charge density

(So )

Find: V at point O

x

y b

dq

r

dx x

Charge/unit length:

3

+

+ + +

+ + + +

+ +

R

dq Total charge Q, uniform

Find: V at centre C

(Homework exercise: review the calculation for the electric field E, which is harder.)

C

+

+ + +

+ + + +

+ +

R

dq Total charge Q, uniform linear density

Find: V at centre C

(Homework exercise: review the calculation for the electric field E)

C

At the center of the semicircle, the potential is: A)  less than kQ/R B)  equal to kQ/R C)  greater than kQ/R

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+

+ + +

+ + + +

+ +

R

r dq

Total charge Q, uniform

Find: V at centre C C

dV (at C) due to dq:

r=R is a constant (same for each dq)

Solution:

1) Find the electric field as a function of r using Gauss’s Law.

2) Imagine pushing a “test charge” in from infinity along a radial line: the potential change with each small change dr in distance is

dV = - E(r) dr.

3) Integrate from R to infinity to find V(R) (relative to infinity) at any position R.

Example: Spherical Charges

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Solid Conducting Sphere,radius R

R r

E

R r

V

Solid Conducting Sphere,radius R

R r

E

R r

V

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R2 R1

+Q

A charge +Q is placed on a spherical conducting shell. What is the potential (relative to infinity) at the centre?

A)  keQ/R1 B)  keQ/R2 C)  keQ/ (R1 - R2) D)  zero

Fields > 3 x 106 V/m will cause a spark in dry air.

What is the maximum potential to which an isolated metal sphere of radius 1 cm can be charged, without causing a spark?

A) 300V

B) 3kV

C) 30kV

D) 300 kV

E) 3MV

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Fields 3 x 106 V/m will cause a spark in dry air. Find the maximum potential on a metal sphere of radius… a) 1 mm

b) 1 m

1)  inside conductor is an equipotential.

3)  surface (just outside.)

4)  Excess charge is on the surface; and

_______________________________________

4) Empty cavity inside a conductor, as well.

5)

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+

-

- - - -

- - +

+ + +

Weak

Large

$ 500

+

Computer chip

Metal foil wrapper

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+Q

A conducting sphere of radius R1, carrying charge Q, is surrounded by a thick conducting shell with no net charge. What is the potential of the inner sphere, relative to infinity?

A)  V = zero B)  0 < V < keQ/R1 C)  V = keQ/R1 D)  V > keQ/R1

-3Q

+Q R1

R2

R3 The dashed green line represents a spherical gaussian surface inside the conducting material. The total electric flux through this surface (in units of Q/ε0) is

A) 0 B) - Q/ε0 C) +Q/ε0 D) -2Q/ε0 E) 3Q/ε0

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So, the charge on the inner surface of the outer shell is

A)  0 B)  - Q C)  +Q D)  -2Q E)  +3Q

-3Q

+Q R1

R2

R3

? +

+ +

++

+

a r r

+

+

Radius a, total charge Q

Uniform volume charge density

Find: V(r) for r<a

answer:

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R3 R2

R1 +Q

-Q

Find: Potential difference

R1 R2 R3

E

r

V

r R1 R2 R3

V1

V2

Find:

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Solution (cont’d):

+

+ + + +

+ + +

+ + +

+ R E = 0

V = constant = Vo

Charge Q

Outside (r > R):

(Just outside)

Vo (on sphere) = V (just outside) =

So…

Quiz:

A large conducting sphere has a net charge Q. A second, smaller conducting sphere with no net charge is now connected to it by a conducting wire. When the system comes to equilibrium, which of the following are true?

A)  the charges on the spheres will be equal B)  the surface charge densities (charge per unit

area) on the surfaces are equal C)  the potentials on the spheres are equal D)  the electric fields just outside the spheres

are equal.

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Solution (Cont’d):

Thus…

Therefore…

i)  Point source:

ii) Several point sources:

or (choose V0 as r )

(Scalar)

iii) Continuous distribution:

OR … I. Find from Gauss’s Law (if possible) II. Integrate, (a “line integral”)

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+

+ + +

+ + + +

+ +

r

r dq

Total charge Q, uniform

Find: V at centre C C

dV (at C) due to dq:

r is a constant (same for each dq)

Solution:

+ + +

++

+

a r r

+

+

Radius a, total charge Q

Uniform volume charge density

Find: V(r)

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Solution (cont’d): Inside (r > a):

Now, Potential: i.e.

So for r > a :

Same as for point charge!!

What about r < a? Can we write

Solution (cont’d):

For r < a:

Note: This is NOT equal to