Download - 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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