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Review 1
Topics
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Electric field, dielectrics and conductors
Problem 21.54
Problem 22.39
Problem 22.70
Problem 23.64
Problem 23.63
Electric field, dielectrics &
conductors
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Electric field
 By definition, it points in the direction of
motion of a positive point charge
Dielectrics
 Molecules are polarized by an external
electric field
Conductors
 In equilibrium, E = 0 within conductor
Problem 21.54
An infinitely long rod of radius R carries a
uniform volume charge density r.
Show that the electric field is
E = r R2/(2e0 r)
E = r r/(2e0)
4
r>R
r<R
…21.54
Gauss’s law  E  dA  Q / e 0 is always true.
However, because this problem has cylindrical
symmetry, the law can be used to find the
electric field within and outside the
rod by using concentric
Gaussian
surfaces
R
r
5
Problem 22.39
Electrons in a TV tube are accelerated from rest
across a potential difference of 25 kV.
Compute the speed with which the electrons hit
the TV screen.
me = 9.1 x 10-31 kg
|e| = 1.6 x 10-19 C
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Hint: conservation
of energy
Problem 22.70
A conducting sphere of radius R1 carries a
charge Q1. It is surrounded by concentric
spherical shell of radius R2 carrying charge
Q2.
Compute the potential at the sphere’s surface,
choosing the potential at infinity to be zero.
7
… 22.70
There are two ways to approach this problem:
1. Compute the potential for the sphere and
the shell separately and add the potentials.
Remember: the potential is always with
respect to some reference value (e.g., zero
at infinity).
2. Compute the potential at the sphere’s
surface directly from
B
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V   E  dr
A
Problem 23.64
A sphere radius R carries a charge Q spread
uniformly over its surface.
Show that the energy stored in its electric field
is
U = k Q2/2R
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… 23.64
This can be done in different ways: e.g., given
the energy density u = e0 E2/2, write
dU = udv, where dv = 4pr2dr is a thin shell of
radius r and thickness dr. Then sum dU.
Note:
Q
Ek 2
r
10
U   udv  
e0
e0  Q 
2
2
k
4
p
r
dr
 2
2 r 

2
1
kQ

2 2 
 4p k Q    
2
 r  R 2R
Problem 23.63
A sphere radius R carries a charge Q spread
uniformly through its volume.
Show that the energy required to assemble the
charge is
U = 3k Q2/5R
while the energy stored within the sphere is
U = kQ2/10R
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Problem 23.63
A sphere radius R carries a charge Q spread
uniformly through its volume.
Show that the energy required to assemble the
charge is
U = 3k Q2/5R
while the energy stored within the sphere is
U = kQ2/10R
12
… 23.63
The total potential energy of the charge is given
U   Vdq
where
q
V k
r
putting together
the pieces we get
13
4 3
and q  r p r
3
2
dq  r 4p r dr
3kQ
U   Vdq 
5R
2
Note: this is the energy
within the sphere as
well as outside