Gauss' Law question about a conducting rod

  • #1

Homework Statement:

In Fig. 23-50, a solid sphere of radius a = 2.00 cm is concentric with a spherical conducting shell of
inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q1 = +5.00 fC; the shell has a net charge q2 = -q1' What is the magnitude of the electric field at radial distances (a) r = 0, (b) r = aI2.00, (c) r = a, (d) r = 1.50a, (e) r = 2.30a, and (f) r = 3.50a? What is the net charge on the (g) inner and (h) outer surface of the shell?

Relevant Equations:

NA
P49.PNG

This is my attempt, i am confused at some points
a. r = 0; The Electric field is 0

b. At r = a/2.00; I verified the answer and it is non zero, but my understanding is that the net charge should be on the surface of the conductor. Hence the charge q1=5*10^-15 C, should go to the surface of the solid sphere. Hence the net charge below "a" should be 0 and the E field should be "0". But why is that the charge does not move to the surface?
If the answer is non zero, then i did the following calculation,
the charge is proportional to the volume hence
q' = q *r^3/R^3;

E= q'/(4πεο*r^2); => E = q*r/(4πεο*R^3); => E = (5*10^-15*9*10^9*a)/(2*a^3)
E= 45*10^-6/(2*4*10^-4) = 5.625*10^-2;

c. and d was easy.
E = q/(4πεο*r^2); r = a for c, r =1.5a for d;

e. r = 2.30a this is the area between the conducting shell, the charge on the inner shell is q2 = -q1. At this r, the net charge is 0 q2+q1=0; E=0;

f. r = 3.50a ; E=0 since the net charge is 0.
g. inner q2 = -q1

h. outer 0.

But my additional thinking let us say that q2 = -2q1;
then on the inner shell the charge will be q2 = -q1 since it has to compensate the solid sphere charge. The remaining charge of -q1 will be on the outer shell. Am I correct? Please advise.
 

Answers and Replies

  • #2
950
418
Hence the charge q1=5*10^-15 C, should go to the surface of the solid sphere. Hence the net charge below "a" should be 0 and the E field should be "0". But why is that the charge does not move to the surface?
The interior sphere is not conductive. They tell you the charge is uniform through the sphere. Also, you will note that in the case of the shell they make a point of telling you explicitly the shell is conductive.
 
  • Like
Likes Physicslearner500039
  • #3
950
418
e. r = 2.30a this is the area between the conducting shell, the charge on the inner shell is q2 = -q1. At this r, the net charge is 0 q2+q1=0; E=0;
Also, that is the interior of a conductor. If the E is not 0. The charges are free to move in the field until it becomes 0. The necessity of the field in the interior of a conductor being 0 is how you arrive at the surface charges, so using them to show the field is zero is a bit of circular logic. I think the statement that the field inside a conductor must be zero is the more fundamental point.
 
  • Like
Likes Physicslearner500039
  • #4
950
418
But my additional thinking let us say that q2 = -2q1;
then on the inner shell the charge will be q2 = -q1 since it has to compensate the solid sphere charge. The remaining charge of -q1 will be on the outer shell. Am I correct? Please advise.
Really, you shouldn’t have to guess at any of this. Between Gauss’ law and the fact that the field inside a conductor is zero you should be very confident of your answer. If you put a Gaussian surface in between the surfaces of the shell the field must be zero so the enclosed charge must be zero, right?
 

Related Threads on Gauss' Law question about a conducting rod

Replies
1
Views
1K
  • Last Post
Replies
5
Views
5K
Replies
8
Views
1K
Replies
4
Views
9K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
6
Views
4K
Top