# Review 1 ```Review 1
Topics






2
Electric field, dielectrics and conductors
Problem 21.54
Problem 22.39
Problem 22.70
Problem 23.64
Problem 23.63
Electric field, dielectrics &amp;
conductors



3
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&gt;R
r&lt;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
6
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
8
V    E  dr
A
Problem 23.64
uniformly over its surface.
Show that the energy stored in its electric field
is
U = k Q2/2R
9
… 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:
E k
Q
r
2
10
U 
 udv  
e0
e0 
Q 
2
k
4
p
r
dr

2 
2  r 


4p k Q  
2

2
2
2

2
1
kQ


r R
2R
Problem 23.63
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
11
Problem 23.63
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   V dq
where
V k
q
and q  r
4
pr
3
3
r
dq  r 4p r dr
2
putting together
the pieces we get
U   Vdq 
13
3 kQ
5R
2
Note: this is the energy
within the sphere as
well as outside
```