E field of a conductor with an arbitrary shape enclosing charge

Click For Summary
The discussion centers on the electric field (E field) around an arbitrary shaped conductor enclosing a charge. It is established that the E field outside the conductor is not radially symmetric, leading to a non-uniform charge density on the interior of a hollow sphere conductor. Applying Gauss's law reveals that the E field in region 4 must be uniform, contradicting the assumption of uniformity in region 2. The participants clarify that assuming uniformity in the E field leads to incorrect results, emphasizing the importance of considering the conductor's geometry. Ultimately, the potential at the conductor's surface must remain constant, raising questions about the accuracy of the answer key.
weezy
Messages
92
Reaction score
5
Screen Shot 2017-06-16 at 3.32.22 PM.png

In this image charge q is enclosed within an arbitrary shaped conductor. I'm asked to find what if the Electric field in regions 1,2,3,4 are uniform and if the surface charge densities on the interior wall of the arbitrary shaped conductor and the hollow sphere conductor are uniform.

My analysis tells me that since the inner conductor is arbitrary the E field outside it won't be radially symmetric and thus should induce a non-uniform charge density on the interior of the hollow sphere. On the other hand since we know E field can't exist inside the hollow conductor, by Gauss's law we must have a uniform field in region 4.

What I am having trouble with is the following fact:

If I am to take a gaussian sphere of radius R in region 2 (it encloses the arbitrary shape) and apply Gauss's theorem, I should find the integral give me ## E = \frac{q}{4 \pi \epsilon R^2} ## which is just the E field produced by point charge located at the origin. However this conflicts the fact that the E field is not symmetric because I expect the arbitrary conductor to have uniform charge density on it's surface but due to it's arbitrary shape it'll have a non-uniform E field in region 2. What is happening here?

EDIT #1: Also I'd like to know if changing the position of q changes the E field in region 2. I think it does not if the arbitrary conductor has some thickness. I'm not entirely sure about this.
 
Last edited:
Physics news on Phys.org
Is the conductor between 1 and 2? If yes, what is 3, another conductor?
weezy said:
My analysis tells me that since the inner conductor is arbitrary the E field outside it won't be radially symmetric and thus should induce a non-uniform charge density on the interior of the hollow sphere.
Right.
weezy said:
On the other hand since we know E field can't exist inside the hollow conductor
That is not true in general, assuming "inside" means vacuum inside the hollow conductor.
weezy said:
by Gauss's law we must have a uniform field in region 4.
That is not inside.

weezy said:
If I am to take a gaussian sphere of radius R in region 2 (it encloses the arbitrary shape) and apply Gauss's theorem, I should find the integral give me ## E = \frac{q}{4 \pi \epsilon R^2} ##
Only if you assume that the electric field has the same magnitude everywhere, otherwise you don't get this result.
Make a wrong assumption and you get a wrong result.
EDIT #1: Also I'd like to know if changing the position of q changes the E field in region 2. I think it does not if the arbitrary conductor has some thickness. I'm not entirely sure about this.
What do you know about the potential of the conductor at its surface? How does that change if you move the charge?
 
  • Like
Likes weezy
mfb said:
Is the conductor between 1 and 2? If yes, what is 3, another conductor?Right.That is not true in general, assuming "inside" means vacuum inside the hollow conductor.That is not inside.

3 is a hollow conductor sphere

mfb said:
Only if you assume that the electric field has the same magnitude everywhere, otherwise you don't get this result.
Make a wrong assumption and you get a wrong result.
Oh my! this is what I overlooked!

mfb said:
What do you know about the potential of the conductor at its surface? How does that change if you move the charge?

The surface must be equipotential so that means in any case the field shouldn't vary. Yet the answer key had my answer marked wrong.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
560
Replies
4
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
23
Views
4K
  • · Replies 1 ·
Replies
1
Views
605
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K