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Each plate of a parallel plate capacitor has a charge $q$ on it. The capacitor is now connected to a battery. Now, which of the following options are correct?

a) The facing surfaces of the capacitor have equal and opposite charges.
b) The two plates of the capacitor have equal and opposite charges.
c) The battery supplies equal and opposite charges to the two plates.
d) The outer surfaces of the plates have equal charges.

Before the battery is connected,

Applying Gauss Law, the inner surfaces of both the plates will have no charge on them, the outer surfaces will have charge $q$.

After the battery is connected,

The potential difference across the capacitor will gradually become equal to that of the battery .The inner surfaces will have equal and opposite charges according to $Q=CV$ where $V$ is the potential difference across the battery.

But, what will be the charges on the outer surfaces of the two plates ?

I am not clear about the role of battery in affecting the charges on the surfaces of the plates.

Option a) Looks right - Not sure about the other options.

Kindly explain what happens to the plates of the capacitor after the battery is connected ?

Does a battery always produce equal amount of negative and positive charges or it may produce additional charges ?

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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.

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  • $\begingroup$ Thank you very much for the response ...I am having a hard time understanding how can charge remain on the outer surfaces while the plates are accumulating equal positive and negative charges on the inner surfaces . $\endgroup$
    – Tanya Sharma
    Commented Jan 17, 2013 at 5:19
  • $\begingroup$ "The important unwavering assumption is a constant electric potential throughout the interior of a plate " How does this help in concluding that the outer charge density remains the same ? "That means that the field on the outside of the plates won't change when connecting the battery" Kindly explain it a little more .I am having diffulty with option d) $\endgroup$
    – Tanya Sharma
    Commented Jan 17, 2013 at 5:31
  • $\begingroup$ @TanyaSharma To argue constant charge on the outside face of the plates: 1. what you quote, that the field inside the plate is zero, 2. the field just outside the plate is the same as before the battery was connected. To transition from some field to zero field, you need a certain amount of charge density on the plane. Since the outside field is the same before/after battery is connected, the charge density on the outside face of the plate must be the same. I argue the outside field is the same b/c 1. total charge is constant 2. distance to charge doesn't affect the field $\endgroup$
    – Alan Rominger
    Commented Jan 17, 2013 at 6:28
  • $\begingroup$ How can we say that the electric field outside the plate is same as before the battery was connected? If we could only prove this then the constant charge density on the outside surface will hold good .Isnt it ?? $\endgroup$
    – Tanya Sharma
    Commented Jan 17, 2013 at 7:03
  • $\begingroup$ @TanyaSharma Yes, and I'm trying to prove it to you. I wrote an equation in my answer that shows that field does not diminish with distance from the charges due to the plate geometry. If our test point is on the right of the plates, then all of the charges are to its left. After you connect the battery they still are. Field diminishes as $1/r^2$ for a point charge, $1/r$ for a line charge, and $1$ for a plate charge. We know that the charges remain inside the metal, and that total charge is the same, so for a test point outside the metal the field is the same. $\endgroup$
    – Alan Rominger
    Commented Jan 17, 2013 at 13:21

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