p361 lecture5a conductors

8/13/2019 P361 Lecture5a Conductors

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Patrick Sibanda


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Conductors in Electrostatic

Equilibrium

The electric field is zero everywhere inside theconductor

Any net charge resides on the conductorssurface

The electric field just outside a chargedconductor is perpendicular to the conductorssurface

By electrostatic equilibrium we mean a situation where

there is no netmotion of charge within the conductor


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Conductors in Electrostatic

Equilibrium

Why is this so?

If there was a field in the conductor the chargeswould accelerate under the action of the field.

The electric field is zero everywhere inside the conductor

++++

++++++++

-------

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

Ein

E E

The charges in the conductor move

creating an internal electric field thatcancels the applied field on the inside ofthe conductor


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The charge on the right is twice the magnitude of the charge on the left(and opposite in sign), so there are twice as many field lines, and they

point towards the charge rather than away from it.


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Combinations of charges. Note that, while the lines are less dense where thefield is weaker, the field is not necessarily zero where there are no lines. In fact,

there is only one point within the figures below where the field is zero can youfind it?


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When electric charges are at rest, the electricfield within a conductor is zero.


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The electric field is always perpendicular to the

surface of a conductor if it werent, thecharges would move along the surface.


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The electric field is stronger where the surface is

more sharply curved.


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Worked Example 4

Any net charge on an isolated conductor must reside on its surface and

the electric field just outside a charged conductor is perpendicular to its

surface (and has magnitude /0). Use Gausss law to show this.

For an arbitrarily shaped conductor we

can draw a Gaussian surface inside the

conductor. Since we have shown that the

electric field inside an isolated conductor

is zero, the field at every point on theGaussian surface must be zero.

From Gausss law we then conclude that the

net charge inside the Gaussian surface is

zero. Since the surface can be made

arbitrarily close to the surface of the

conductor, ny net ch rgemust reside on theconductors surface.

!

E!!! d!

A=Q

in

!0


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Worked Example 4 contd

We can also use Gausss law to determine the electric field just outside

the surface of a charged conductor. Assume the surface charge density

is . Since the field inside the conductor is

zero there is no flux through the face

of the cylinder inside the conductor. If

had a component along the surface

of the conductor then the free charges

would move under the action of thefield creating surface currents. Thus

is perpendicular to the conductors

surface, and the flux through the

cylindrical surface must be zero.

Consequently the net flux through thecylinder is EAand Gausss law gives:

0 0 0

orin

E

Q AEA E

! !

" " "

# = = = =


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Worked Example 5

A conducting spherical shell of inner radius aand outer radius bwith a

net charge -Q is centered on point charge +2Q. Use Gausss law to

find the electric field everywhere, and to determine the chargedistribution on the spherical shell.

a

b

-QFirst find the field for 0 < r< a

This is the same as Ex. 2 and is the field due to a

point charge with charge +2Q.

2

2

e

QE k

r=

Now find the field for a< r< b

The field must be zero inside a conductor in equilibrium. Thus fromGausss law Qin is zero. There is a + 2Q from the point charge so we

must have Qa= -2Qon the inner surface of the spherical shell. Since the

net charge on the shell is -Qwe can get the charge on the outer surface

from Qnet= Qa+ Qb.

Qb= Qnet- Qa= -Q - (-2Q) = + Q.

+2Q


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Worked Example 5 contd

a

b

-Q

+2Q

Find the field forr>

b

From the symmetry of the problem, the field

in this region is radial and everywhere

perpendicular to the spherical Gaussian

surface. Furthermore, the field has the same

value at every point on the Gaussian surfaceso the solution then proceeds exactly as in

Ex. 2, but Qin=2Q-Q.

( )24E dA E dA E dA E r!" = = =" " "Gausss law now gives:

( )2 2 20 0 0 0

2 14 or

4

ine

Q Q Q Q Q QE r E k

r r!

" " " !"

#

= = = = =


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p361 lecture5a conductors

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