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Some question on charge inside cavity

  1. Mar 5, 2012 #1

    Consider the case of a charge placed inside the cavity in a metal. Let the charge be at A. Assume that the cavity is some weird geometric shape so that one point on the inner surface of the cavity is closer than some other point. Let point B on inside wall be closer to the location of charge,( that is A) than the point C, which is farther. I was reading some thread at PF, which asked where the density of the induced charge will be greater. Of course the answer is B. But following reasoning was offered by one poster for this. Since both points B and C are on the wall of conductor, they are at same potential. So the potential difference between A and B would be same as potential difference between A and C. But since potential difference between the two points is basically an integral of E over the path (from A to B or from A to C), and since the path between A and C is more than the path between A and B, it should follow that the E at B will be more since it has shorter path. The potential difference between A and B is same as that between A and C. So shorter path should lead to greater electric field.

    Now I have some questions about this reasoning. We can only conclude that the
    E will be more along the path A-> B. How can we conclude that the E will be greater
    at point B ?

  2. jcsd
  3. Mar 5, 2012 #2
    In fact E is zero inside the cavity. Since its zero, its integrals along
    both paths are zero and of course equal. Within a cavity inside a conductor, Electric field is zero, as if it there was no cavity there.
  4. Mar 5, 2012 #3
    Hassan, I said there is charge inside the cavity. How does E vanish inside the cavity ?
  5. Mar 5, 2012 #4
    You seem to have several misconceptions running together, and I can't pin point which is the problem in this case. The electric field is the negative gradient of the potential, which is the three-dimensional slope. Think of the potential as a mountain terrain and the electric field as a field of arrows covering the mountain terrain, at each spot pointing in the direction where the mountain is steepest. A constant potential (equipotential) would be like the ring of points on a mountain that are all at the same altitude. These points do not necessarily all have the same slope (the same electric field), but the electric field vector does point perpendicular to the equipotential line. So, the internal edge of your cavity would be like the mountain's base, and the charge in the cavity would be like the mountain's peak. Picture a point charge in a spherical cavity, offset to the right of the center. In order for the mountain terrain to reach from the peak to the base (charge to cavity wall) on the right side across a shorter distance, the slope (the electric field) must be higher.

    So: shorter path but stronger E field cancel out to give same potential difference between charge and wall.
  6. Mar 5, 2012 #5
    Chris, I actually realised that I wanted to ask question about the charge densities at B and C. But the explanation given by the poster (in some thread on PF) in my post actually does answer it. Since there is stronger E near the point B, there is greater charge accumulation.
    I am embarrassed that I asked the question about the E at B. That was a mistake. So all is well ....... Thanks
  7. Mar 5, 2012 #6
    Please forgive me for misreading your question, and underestimating it this way. I should be sleeping at this hour.

    As for your question, I'm with you. We can imagine two decreasing functions starting with the same value. One of them first dropping rapidly and then at distant points approaching a nonzero value. The other drops less rapidly but then goes to zero ( or any value less than the final value of the previous function) and at a nearer point. The areas under the curves could represent the line integrals. Both could be the same while the one with the shorter distance have a lower final value. Here we need to resort another reasoning to prove that Electric field doesn't drop like that.
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