- #1

issacnewton

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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 ?

thanks