Conducting Sphere With Cavity E-field / Gauss' law

  • #1

Homework Statement


You have a conducting sphere that is in equilibrium, it has a cavity in it with positive charge +q. If you bring another charge +q2 near the outer edge of the conductor does the total surface charge on the wall of the cavity, q(int) change? There is an image attached that might explain the situation better.

Homework Equations


Not sure, possibly just logic

The Attempt at a Solution


The issue that I am having is when you bring q2 close to the conducting sphere, my intuition tells me that the charge on the wall of the cavity q(int) should redistribute throughout the conductor, leaving no excess charge anywhere on the conductor? However, the correct answer is the surface charge on the wall of the cavity q(int) doesn't change. Any help would be much appreciated!
Thanks
 

Attachments

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,989
3,142
Draw a Gaussian surface entirely inside the conducting material such that it encloses the entire cavity.
Please set your intuition aside and answer the following four questions in sequence. Then you will see what's going on. It's a matter of logic, not intuition.

1. What is the integral ##\int_S \vec E \cdot \hat n~ dA~##?
2. What does your answer to the previous question imply about the charge enclosed by the Gaussian surface?
3. What do your answers in the previous two questions imply about the charge on the inner surface of the cavity?
4. How will your answers in the previous three questions change (consider them in sequence) if some other charge is brought near the outer surface of the conductor and why?
 
  • Like
Likes cookiemnstr510510
  • #3
Draw a Gaussian surface entirely inside the conducting material such that it encloses the entire cavity.
Please set your intuition aside and answer the following four questions in sequence. Then you will see what's going on. It's a matter of logic, not intuition.

1. What is the integral ##\int_S \vec E \cdot \hat n~ dA~##?
2. What does your answer to the previous question imply about the charge enclosed by the Gaussian surface?
3. What do your answers in the previous two questions imply about the charge on the inner surface of the cavity?
4. How will your answers in the previous three questions change (consider them in sequence) if some other charge is brought near the outer surface of the conductor and why?
Ahhh, okay:
1. 0
2. that the charge enclosed by the guassian surface sums to zero
3. the charge on the inner surface of the cavity must be -q
4. They will not change
This makes a lot of sense
 
  • Like
Likes TSny
  • #4
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,989
3,142
Ahhh, okay:
1. 0
2. that the charge enclosed by the guassian surface sums to zero
3. the charge on the inner surface of the cavity must be -q
4. They will not change
This makes a lot of sense
Now that you reasoned it out, think of it qualitatively this way.
1. The only way that charges can communicate their presence to other charges and be affected by them is through electric fields.
2. Under static conditions, the electric field inside the conductor is and remains zero.
3. Therefore any charge on the inner surface of the conductor cannot be affected by any charge outside the conductor and vice-versa.
 
  • Like
Likes cookiemnstr510510
  • #5
Now that you reasoned it out, think of it qualitatively this way.
1. The only way that charges can communicate their presence to other charges and be affected by them is through electric fields.
So lets say there were no charge within the cavity. We know that the conductor was in equilibrium, we also know that to be a conductor it has to have one loosely bound electron per atom. So the charges within the conductor would have no way to interact/communicate since they were not in an electric field? Is this what you're saying?
 
  • #6
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,989
3,142
So lets say there were no charge within the cavity. We know that the conductor was in equilibrium, we also know that to be a conductor it has to have one loosely bound electron per atom. So the charges within the conductor would have no way to interact/communicate since they were not in an electric field? Is this what you're saying?
Not exactly what I am saying. If you bring an external charge near an uncharged conductor, there will be a transient instantaneous electric field inside the conductor that will cause the free electrons inside the conductor to move. They will keep on moving until they have no reason to do so, i.e. when the electric field inside the conductor is zero and the static condition is reached. The approach to equilibrium is very fast and depends on the conductivity of the conductor. What I have previously described is the bottom line. When all is said and done and the transients have died out, it's as if only the outer, but not the inner, surface charge distribution changes.

You raised a good point.
 
  • Like
Likes cookiemnstr510510
  • #7
When all is said and done and the transients have died out, it's as if only the outer, but not the inner, surface charge distribution changes.

You raised a good point.
The outer meaning the the side closest to the external charge?
 
  • #8
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,989
3,142
The outer meaning the the side closest to the external charge?
Yes, the external surface that forms the boundary between the conductor and the vacuum outside it.
 
  • Like
Likes cookiemnstr510510

Related Threads on Conducting Sphere With Cavity E-field / Gauss' law

Replies
3
Views
3K
Replies
16
Views
11K
  • Last Post
Replies
4
Views
172
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
723
Top