Ph 213 – Challenging Problems (set 4)
Due: July 23, 2013
Name: _______________________
1. Consider a very long (infinite) charged cylinder of radius R with a non-uniform charge
density inside it that varies linearly according to (r) = Br, (r < R) where r is the distance
from the central axis and B is a
constant.
A) Calculate the electric field
everywhere (inside and outside
of the cylinder). Show your
work, and diagrams.
Due to charge distribution there are two regions for Electric field, inside of the cylinder
(r < R) and outside of it (r > R). Use
Gauss’s Law separately for each region.
Region I – E-field inside cylinder (r < R)
We choose a pillbox of radius r and
length l as Gaussian surface:
Electric field points away from the central axis. The electric flux through our Gaussian
surface is:
  Qenc
 E  dA 
0
Area vector dA is away from central axis from body of Gaussian surface and is parallel to E, bud
dA is outward from the caps hence it is perpendicular to E field.


 E  dA  ( EdA cos 90
0  E I 2 r l  0 
o
) left cap  ( EdA cos 0) body  ( EdA cos 90 o ) right cap 
Qenc
0
Qenc
0
The charge enclosed is the amount in the Gaussian surface of radius r. To find the enclosed
charge when choose a cylindrical shell of radius r, length l, and thickness dr that has volume dV.
r
2 3
Qenc   dV   Br (2rldr ) 
Br l
3
0

Qenc 2Br 3l
B 2 
E I 2 r l 

 EI 
r r
0
3 0
3 0
In region II (r > R) a new pillbox of radius r and length
l as shown is chosen for Gaussian surface.
Electric field points away from the central axis. The
electric flux through our Gaussian surface is:
  Qenc
E
  dA 
0
Area vector dA is away from central axis from body of Gaussian surface and is parallel to E, bud
dA is outward from the caps hence it is perpendicular to E field.
All steps and integration is same as above except the enclosed charge will be entire charge on
cylinder:
R
2
Qenc   dV   Br (2rldr ) 
BR 3l
3
0

Q
2BR 3l
BR 3 1 
E II 2 r l  enc 
 E II 
r
0
3 0
3 0 r
Here is the summary of the electric field inside and outside of the cylinder:
B 2 

rR
 E I  3 r r

0

3
 E  BR 1 r
rR
 II
3 0 r
B) Calculate the electric potential everywhere outside of the cylinder.
We can Choose the zero reference point for the potential at r = 0, i.e V(0) = 0
For region II (r > R) we are taking a path form the central axis (r = 0) radially through regions I
and regions II and so we need to use both functional forms for the electric field in the appropriate
regions. The potential difference between any point lying on a circle of radius r > R and the
central axis is given by the line integral expression
R 
 r 

V (r )  V (0)   E I  dl   E II  dl (dl and E are parallel)
0
R
R
r
B 2
BR 3 1
r dr cos 0  
dr cos 0
3 0
3 0 r
0
R
V (r )  V (0)   
V (r )  V (0)  
BR 3 BR 3 r

ln
9 0
3 0
R
C) A point charge of mass m and charge q has been held at point P a distance of 1.5R as
shown. Find the speed of the point charge when it reaches a distance of 3R from the
central axis of the cylinder.
Since the charge is positive, therefore if released from rest it will accelerate in the direction of
electric field, by converting electric potential energy to KE. The change in kinetic energy (KE)
when the object moves from a distance r = 1.5R from the central axis of the cylinder to a distance
r = 3R is given by
KEf – KEi = –(Uf – Ui) = –q(Vf – Vi)
Recall we chose V(0) = 0 at central axis of cylinder (r = 0), so:
R 
 r 

V (r )  V (0)    E I  dl   E II  dl (dl and E are parallel)
0
R
Vi(r = 1.5R) = V(1.5R) – V(0) and Vf (r = 3R) = V(3R) – V(0)
Therefore Vf – Vi =(3R) – V(1.5R)
3R 
3

3 R BR 1
V  V f (3R)  Vi (1.5R)    E II  dl  
dr cos 0 (dl and E are parallel)
1.5 R 3
r
0
1.5 R
BR 3
V  
(ln 3R  ln 1.5R)
3 0
KEf – KEi = –qV
vf 

BR 3
 V  
ln 2
3 0
1 2
BR 3
mv2  0  q(
ln 2)
2
3 0
2qBR 3
ln 2
3m 0
2. On the x axis at distances of x = ± d and x = 0 three infinite vertical sheets of charge are
located. The sheets have uniform charge densities – andas shown.
A) Calculate the electric field at each region i – iv shown.
Let’s choose a pillbox as the Gaussian
cylinder with end caps in regions (i) and
(iv) as shown. The total charge enclosed
is zero and hence the electric flux on the
end caps must be zero. Thus the electric
field must be zero in regions (i) and (iv).


 E  dA 
Qenc
0
 (  2   )( area )  0
Thus the electric field must be zero in regions
(i) and (iv).
For regions (ii) and (iii), we choose a Gaussian
pillbox (cylinder) with end caps in regions (ii)
and (iii), as shown.
  Qenc
 E  dA   0
Where Qenc = (+2)A with A being the area of
the pillbox.


 E  dA  ( EdA cos 0
EA  EA 
2A
0
o
) left cap  ( EdA cos 90) body  ( EdA cos 0 o ) right cap 
 E
2A
0

0

 
So taking Gaussian surface extended between region (j) and (ii) will yield to E   i in
0



region (ii). In same way it will be E   i for region (iii).
0
The summary of electric field for each region is:
 x  d region (i )
0
  
 i
  d  x  0 region (ii )

 0
E ( x)  
  i
 0  x   d region (iii )
 0

 x   d region (iv )
0
B) Plot the x component of the electric field, Ex as a function of x. Clearly indicate on the
vertical axis the values of Ex for the different regions.
The graph of the x-component of the electric
field, Ex vs x is based on above field in each
region:
C) Find the electric potential in each of the
four region labeled i – iv as shown (take
V = 0 when x = ∞)
The electric potential difference between infinity and any point P located a distance x from
origin, is given by
x 

V ( x)  V ()    E  dl (area under the graph of E vs x)

Now we use area under graph of E to evaluate the integral for each region. First for a charge
brought from infinity to point P where x > d
x



In region (iv), E = 0, hence V ( x)  V ()   E  dl  0

Viv ( x)  0

For region (iii), the particle has to go through region (iv) first (0 < x <+d
x 



V ( x)  V ()   Eiv  dl  ( Eiii  dl )
d

d
The potential is the negative of all area under graph of E and x axis
(from ∞ to shaded area at x):

Viii ( x)  V ()  0  ( x  d )
0


d x
0
0
For region (ii), the particle has to go through region (iii) and (iv) first coming from infinity.
d 
0 
x 



Vii ( x)  V ()   Eiv  dl  ( Eiii  dl )  (  Eii  dl )
Viii ( x) 

d
0
The potential is the negative of all area under graph of E and x
axis (from ∞ to shaded area at –x):
Vii ( x)  o  (


(0  d ))  (( (0  ( x)))
0
0


x d
0
0
Finally for region (i), the particle has to go through region (iv), (iii), and (ii) first coming from
infinity:
d 
0 
d 
x 




Vi ( x)  V ()   Eiv  dl  ( Eiii  dl )  (  Eii  dl )  (  Ei  dl )
Vii ( x) 

d
0
d
The potential is the negative of all area under graph of E and x axis (from ∞ to x < –d):
Vi ( x)  o  (


(0  d ))  ( (0  (d )))  (0)
0
0
Vi ( x)  0
Following is the summary of potential for all regions:
 x  d region (i )
0

 d x
  d  x  0 region (ii )
0
 0
V ( x)  
 d   x
 0  x   d region (iii )
 0
0

 x   d region (iv )
0
D) Plot the electric potential as a
function of x. Indicate units on the
vertical axis.
The graph at right shows the plot of
the electric potential as a function of x
with units of [volts] on the vertical
axis.
E) How much work is done to bring a
point charge of +Q from infinity to
the origin?
The work one must do to bring a charge (point-like charge) is equal to the change in potential
energy (assuming the charge begins and ends at rest). Therefore
W  U 0  U   Q(V0  V )
Q
W 
d
0