# Induced charge in a conductor

My question here is based on the following question in Physics for Scientists and Engineers (3rd edition) by Randall Knight:

An electroscope is positively charged by touching it with a positive glass rod. The electroscope leaves spread apart and the glass rod is removed. Then a negatively charged plastic rod is brought close to the top of the electroscope, but it doesn’t touch. What happens to the leaves?

a. The leaves get closer together.

b. The leaves spread farther apart.

c. One leaf moves higher, the other lower.

d. The leaves don’t move.

The answer given is a. Let's say that the total charge on the electroscope is +Q. I'm guessing that as you bring the negatively charged rod near to the electroscope, the positive charge on the electroscope "shifts" towards the negative rod, as follows: But it seems to me that there could (in theory?) be other possible configurations of the total charge, such as this: Here, the total charge on the electroscope is still +Q, but one end has charge +2Q and the other end has charge -Q. This would actually change the answer to the question (since the leaves would be further apart), but I'm more interested in the underlying physics than the answer to the question.

My question is simply this: how do we know (or, more accurately, how am I supposed to know) that the configuration of charge in the above image cannot occur?

Another question I have (on the exact same topic) is the following. If you take a spherical conducting shell (or basically any conductor with a cavity) and place a charge +Q in the cavity, then I know that the induced charge on the inner surface has to be -Q (by Gauss' law), as follows:

(Here I'm not worrying about the charge on the outer surface). But now my question is this: how do we know that the charge distribution on the inner surface is all negative? Couldn't there be an excess of negative charge on the part of the surface closest to the charge in the cavity, leaving a positive charge on the opposite side of the inner surface, while keeping the total charge on the inner surface -Q? That is, couldn't something like this happen: And if this is not possible, why not?

I guess the common theme underlying these two questions (as well as the previous one I asked) is that I have no idea how to "intuitively" work out how a charge will distribute itself over a conductor. And it seems intuitively that there could be multiple possible distributions and it's not obvious (to me) which is "correct". The books I'm looking at don't seem to have many guidelines about this.

Your first question. You are right if the charge on the negative rod is large enough. As you bring the rod closer to the cap of the positively charged electroscope, the leaves will first go down, as free electrons are forced down into the leaves, neutralising their positive charge. As you move the rod closer still, more electrons are forced into the leaves, so they now have a negative charge and will repel each other (as long as the negative rod is kept in place).

Note that the central assembly (top cap, stem and leaves) are made of metal, so the only mobile charges are electrons (negative). It's best not to talk in terms of positive charges moving, though the charge distribution can sometimes change as if positive charge had moved.

Your second (harder) question. My attempt at an answer is based on the principle that in an electrostatic set-up, field line bundles start and finish on equal and opposite charges.

We start by postulating that the charge distribution is as shown on your lower diagram. Consider a gaussian surface inside the metal, a surface that surrounds the cavity and therefore contains the isolated charge, $$+Q,$$ and the surface charges on the cavity walls. No flux can go through the gaussian surface because $$\vec E=0$$ everywhere on the surface. Therefore the total charge enclosed by the surface is zero, so if the charge induced on the cavity wall near $$+Q$$ is $$-Q',$$ the positive charge on the 'far' wall of the cavity must be $$Q'-Q$$ in which $$Q'>Q.$$

Now consider another gaussian surface, containing the isolated $$+Q$$ and the negative charge, $$-Q'$$ on the cavity wall. Net flux of magnitude $$\frac{1}{\epsilon_0}(Q'-Q)$$ must enter this surface, and can do so only through the part of the surface that runs through the cavity. This flux must originate from the positive charges, $$Q'-Q,$$ on your opposite cavity wall.

We therefore have field lines crossing the cavity from the positively charged area of wall to the negatively charged area. So work would have to be done to take a positive test charge from the latter area to the former. This implies that there is a potential difference between the two patches of wall. But this can't be the case as this would cause a current through the metal, and there isn't one, as we have postulated an electrostatic situation! Thus we can have no positive charges on the cavity wall.

• So would you agree that the question in the book is fundamentally ambiguous? As for the second question: I'm not sure I'm convinced (or fully understand) your answer. Though it could well be correct. – Jesse Mar 5 at 0:26
• I believe that it is ambiguous. Do try and pin down what you don't understand about my answer to your second question, and I'll try and explain, but I shan't be able to do so for a few hours. Starting point: we agree, I hope, that the algebraic sum of positive and negative charge on the cavity wall is equal to the isolated positive charge placed in the cavity. – Philip Wood Mar 5 at 7:58
• I think what I don't understand is the sentence starting with "That implies..." In particular I can't see why the previous sentence implies what you say (that there's a potential difference). I guess this is based off the formula $\int \mathbf{E} \cdot dl = V(a) - V(b)$, but I'm not sure. (As for the charge on the cavity wall: do you mean that the total charge is -Q?) – Jesse Mar 5 at 19:03
• (1) "As for the charge on the cavity wall: do you mean that the total charge is -Q?" Yes, the algebraic sum of negative and positive charge. This implies that the actual negative charge on the wall is greater (more negative) than -$Q$. (2) If some field lines run from the positive charge on the wall to the negative (as I believe they do) then work would be done on a (positive) test charge going from one to the other, so by definition there is a p.d between these areas of wall. But there can't be, or there'd be a current through the conducting wall! – Philip Wood Mar 5 at 19:45
• "If some field lines run from the positive charge on the wall to the negative ... then work would be done on a (positive) test charge going from one to the other". Yes, I think I agree with that now. To put it more mathematically, the potential difference would be $V(a) - V(b) = \int_a^b \mathbf{E} . dl$, which is positive because the integral on the rhs must be positive. But the potential must be the same everywhere in a conductor, so this is a contradiction. But now it's not clear to me why there must be field lines going from the positive charge on the wall to the negative charge...! – Jesse Mar 6 at 0:34