when a capacitor is charged by connecting to a battery having plates of different areas, will the charges on the two plates have equal magnitude, or may they be different?

Question

in my tentative answer is they will be different, since Q = CV and capacitance is directly proportional to the plate area so they will have different charges. but others say that only the surface charge densities of the plates will be different but the magnitude of charge on both plates will remain the same. I am so confused can someone help me with this. It would mean a lot.

Answer 1

Great question!

Short answer - the charges will be the same in magnitude because the battery is just a pump - it can only move charges from one plate to the other.

There are two important points here:

1) charging a capacitor

2) geometry of a capacitor

Let's start with point #2.

Yes, in principle, you can have a capacitor that have different charges on each plate. You can think about the charge on plate 1 as Q/2+q and the charge on plate 2 as Q/2-q, where Q is the total (overall) charge of the entire capacitor and q is relative charge between the plate. The latter is what we would typically deal with in a capacitor problem. If you play around with combinations Q/2+q and Q/2-q you can realize that they can produce any numbers you like independently from one another.

Now back to point #1.

(a) If you connect both plates to each other and to some metallic object that already has some charge, that charge can move around redistributing itself between that object and plates of the capacitor. The actual value of charge on each plate and the object will depend on the geometry. Think about a simple extreme example: You take a small charged metallic ball and place it in the center inside two concentric spheres, one slightly larger then the other. If you connect everybody with a short wire in a most direct fashion (along the radius) the charge will simply distribute itself over the largest (outer) shell. This is because elementary charges there will be as far from each other as possible - after all they are charges of the same sign and, thus, repel each other.

(b) This is NOT what we do in the problem. We charge that capacitor with a battery. Now, what does a battery do? It takes (+) charge from its negative terminal and deposits it on the positive terminal for you to use it. This is done by chemical reactions inside the battery. In other words, it can not create charges - it moves them, like a pump. Now if we connect battery to a capacitor, the battery will take some charge from one plate and move it to the other. The system will remain overall neutral at all times. Thus the plates will acquire equal and opposite charges irrespective of the geometry of the capacitor.

Now, diving a lot deeper:

When we charge that capacitor using a battery we do it with wires. Well how else you can do it? The thing is that wires form a capacitor of their own, and so you are really charging both - the target capacitor and the one made of the wires that you are using. They are parallel to each other. Now imagine (drawing it will help a lot) we exaggerate a bit and make wires thick flat and wide, and palace them near each other to form a parallel plate capacitor. Let say the original capacitor is also a parallel plate capacitor of the same thickness. Let's put them next to each other, such that one plate of the wire capacitor partially hangs over the opposite plate of the original capacitor. Let's trim this combination from sides to produce a nice parallel plate capacitor overall. The plates of that overall capacitor will acquire charges +q and -q as usual, distributed uniformly throughout each (combined) plate. Yet because areas of segments that make each plate are different, they will have different charge. When you disconnect these capacitors, each plate will be charged differently depending on the area and where they are located (electric field). Finally, removing our wires-on-steroids capacitor, we will end up having that original capacitor with different charges on each plate - different in magnitude. That is, the capacitor will be overall charged with non-zero Q. The wires will end up being overall charged with -Q, because the system must still remain overall neutral - remember that battery is just a pump. In reality this effect will be minimal because the capacitance of the wires is usually very very small.