Download Physics PHYS 354 Electricity and Magnetism II Problem Set #3

Physics PHYS 354
Electricity and Magnetism II
Problem Set #3
1.
Conducting Sphere in a Uniform Electric Field
A conducting sphere of radius R carrying a charge +Q is placed in a uniform
electric field of intensity E0 . Show through application of Laplace's equation in
spherical coordinates that the electrostatic potential V r,  at a given point
outside the sphere is given by
V r,   E0r cos  
E0 R 3
cos  .
4 0r
r2
Q

Could you have predicted this result from the result of Problem 1 on Problem Set
#1? Why or why not? From this expression determine    on the surface of the
sphere. Also show that the total charge on the surface of the sphere is +Q.
Finally, determine the dipole moment of the induced charge on the sphere.
2.
Cavity in a Dielectric
A large block of
dielectric material (


D  E ) is uniformly
polarized with
  
E x  E0 . A small
spherical cavity of
radius a is cut out of the
block, as shown in the
figure below.
E0
a
z

Calculate the field E inside of the cavity:
a)
 
assuming that P  x  adjusts to the presence of the cavity, so that in r a ,
 
 
 
D E , in r a , D E , and as r  , E E0 . (Hint: Choose suitable
forms for  in r a and , then match Dr and E at r a .)
b)
 
 
assuming P  x  is frozen at its original value in r a and P 0 in r a .

Find the electric field E by two methods:
i)
ii)
3.

Subtract the field due to an oppositely polarized sphere from E0 .
 


Calculate it by replacing P with    P and  P nˆ and
finding appropriate  in r a and r a .

 
Lock in the polarization by using the fact that D  0 E  P0 where

 
P0 is the "locked in" polarization, rather than D E , where the

polarization depends on E .
Laplace's Equation in Plane Polar Coordinates
In this problem you will be asked to first explore Laplace's equation in plane polar
coordinates, then solve a particular problem by matching the boundary conditions.
a)
Take Laplace's equation in cylindrical coordinates ( r, ,z ) and convert it
into an equation in polar coordinates ( r, ). Assume the solution has the
form
V r,  Rr   .
Use a separation parameter k 2 to show that the solution is
V (r, )  A0 B0 ln r C0 D0 
V (r, ) Ak r Bk r
k
b)
k
C
k
cos k Dk sin k 
when k 0 , and
when k 0 .
Consider a wedge-shaped region bounded by two grounded conduction
surfaces intersecting at the origin with an interior angle  together with a
line charge of strength  per unit length located at the point r0 ,  within
the wedge, as shown below.
2
V=0



r

VI
r0
V=0
P(r,)

VII
Show that the potential in the region bounded by the intersecting
conducting planes is
n
 n   n
  1 r  
  sin
VI r  r0  
sin

 0 n 1 n  r0 
    

 and

n
 n   n
  1  r0  
VII r  r0  
sin
  sin

 0 n 1 n  r 
    



where it is apparent that the solution must be written in two parts, VI for
the region r r0 , and VII for the region r r . The electric field given by
the two solutions must follow the correct boundary conditions when r r0 .
(Hint: In polar coordinates, the sampling property of the delta function is
written,
b
r0  f     d  f   where 0   ,
0
and the line charge density can be written:        .)
3