I read in a book that if both the plates of a parallel plate capacitor are given equal positive charges $q$, then the charges on the facing surfaces will be zero and the charge on the outer surfaces is $q$ for each plate. How is this possible? Shouldn't the charge distribute uniformly on both the surfaces of each plate?
[Physics] Two capacitor plates with equal positive charges $q$
capacitanceconductorselectrostaticsgauss-law
Related Solutions
I think you question can be answered succinctly from this point:
Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?
A battery never produces additional charge. Very few things can do this (like a Van de Graaff generator), and even those can only do so locally. A battery pumps charge, and that leaves no room for adding to the total charge count (c).
I thought it strange that the problem is starting out saying that the capacitor has on net extra charge, but you have analyzed the situation correctly. Since we're treating it as an infinite parallel plate capacitor, before the battery is connected the charge is concentrated on the outside face of both plates.
When the battery is connected, and the capacitor has fully charged (if you neglect internal resistance this last specifier isn't needed), then you will have additional positive charge on one plate and additional negative charge on the other. Following the eqi-potential principle, these opposite charges will gather on the inner faces of the two plates (a).
Now a deeper question: will connecting the battery affect the charge density on the outer faces of the plates? I will argue "no". The important unwavering assumption is a constant electric potential throughout the interior of a plate. Introducing the battery introduces a field in-between the inner surfaces, which results in potential difference equal to the voltage of the battery. You have 4 points (which are infinite planes) along the number line (since this is 1D symmetry) where charge is located, and the field is $E=\sigma/(2 \epsilon_0)$ pointing away from the plane - that is, proportional to the charge density and constant. That means that the field on the outside of the plates won't change when connecting the battery, since the total charge quantity stays the same, the field doesn't diminish with distance from the charges, and you only moved charges around. The field within a plate is zero and the outside charge density remains the same, thus, in order to transition from a zero field to the unchanged outside field strength, you require the same charge density on the outside face (d). The potential on the surface is different, but the charge is the same - a very important nuance of electronics.
Thus, options "a", "c", and "d" are both correct.
By Gauss's theorem, we have (Let $S$ be an area of a capacitor):
charge $q$ creates an electric field $\lvert E \rvert = q/2\epsilon_0 S$ upside and downside.
I will use your example in Edit. Let me call a plate with $2Q$ plate A, and the other plate B. Let a charge in downside of plate A be $q_1$, and upside of B $q_2$. Hence chrge on upside of A is $2Q - q_1$, and downside of B is $-Q-q_2$.
Now, consider the point inside of plate A. There is $2Q- q_1$ above, and $q_1 + q_2 + (-q_2 - Q) = q_1 - Q$ below. Thus, electric field inside plate A is $(3Q- 2q_1)/\epsilon_0 S$. Since this must be $E = 0$ (by induction!), $q_1 = 1.5Q$.
By doing the same in plate B, we have $q_2 = -1.5Q$.
Finally, consider an electric field between plates. Since, upside of A and downside of B (namely, outside of capacitor) have same charge $Q/2$, they don't create electric field. Charges inside of capacitor (downside of A and upside of B) create electric filed $E= 1.5Q/\epsilon S$. (Hence we have $1.5Q = C \times Ed$)
In general cases, let charge on A be $Q_1$, and charge on B be $Q_2$. Then, a charge upside of A and downside of B (outside of a capacitor!) are both $(Q_1 + Q_2)/2$, so they don't create an electric field inside the capacitor. We only need to consider charges inside the capacitor (downside of A and upside of B), which are $\pm(Q_1-Q_2)/2$. In this case, $$ \frac{Q_1 - Q_2}{2} = CV . $$
Best Answer
In a perfect the conductor the charges are free to move wherever they "want". Where they "want" to be is as far away as possible from other charges of the same sign (positive, in this case). After all, like charges repel each other. Consequently, they move to the outer surfaces of the plates, maximizing the distance between them.
Here's another way of saying the same thing: the charges are moving in order to minimize the total potential energy of the system.
Note that that is not the whole story, because it doesn't quite explain why the charges have to be right on the surface. Couldn't potential energy be minimized by some distribution that leaves some charges within the bulk? The answer is no, this cannot happen. A perfect conductor cannot maintain an electric field inside the bulk of the conductor. Otherwise charges within the bulk would feel this field and move in response. The only steady state condition that can be reached is the one where there is no electric field in the bulk.
In the parallel plate capacitor, the field due to the charges on the left side of the left plate exactly cancels the field due to the charges on the right side of the right place, with the result that the field in both conductors is zero (and all the charges are on the outside surfaces).