# A Cavity Within a Conductor

1. Aug 4, 2011

### sparkle123

This is an excerpt from my textbook. Could someone please help me understand why the line I highlighted in yellow is true? Thanks! :)

2. Aug 4, 2011

### kuruman

Suppose E is non-zero and in fact looks as shown in the drawing. Then if you chop the path along the line from A to B in many, say 1000, small pieces ds, you can calculate 1000 dot products E.ds each of which is equal to Eds and positive because the cosine of the angle between E and ds is +1. Now if you add 1000 positive numbers, what do you end up with?

3. Aug 4, 2011

### mathfeel

I don't think one can assume that this path drawn on the textbook necessarily follows some unknown field line. Along the curve, the E field might not might not be parallel to dS.

But we can improvise: start from A, follows the field line like you suggested to some other point C, then go along the cavity surface until you reach B. In the A-C leg of the curve, the potential different would be non-zero if there is a finite field. The C-B leg of the curve contributes nothing to the potential different since electric field is always normal at a conductor surface. Note that A-C cannot be done if there is some point charge inside the cavity...

Anyway, I think this particular argument of the textbook is not very good. What if there is a some charge in the cavity? In this case, electric field is clearly not zero. Yet the potential difference between A and B is still zero.

4. Aug 4, 2011

### kuruman

The statement is "... we can always find a path ..." I have found such a path. It is the one in which E is always parallel to ds.
If there is a charge, yes you are right. However, the textbook clearly states "Let assume that no charges are inside the cavity." You cannot change the assumptions then claim that the argument is no good.

5. Aug 4, 2011

### sparkle123

thanks to both! but what if the field was entirely contained within the cavity, like a loop or something?

6. Aug 4, 2011

### kuruman

Static electric field lines (as might be the case here) start at positive charges and end at negative charges. If they formed closed loops, the integral $\oint \vec{E}\cdot d \vec{s}$ would not be zero contradicting the conservative nature of a static electric field.

7. Aug 4, 2011

### sparkle123

oh okay thanks! :)