Question

Suppose a charge'+Q1' is given to positive plate of capacitor and a charge'-Q2' to the negative plate of capacitor.what is the "charge on the capacitor"??

Answer 1

The net charge on the cap is
\[Q_{net}=Q_{1}-Q_{2}\]

Answer 2

No it isn't. It's just Q. If your capacitor is charged up so that 1 plate has charge +Q and the other has charge -Q, then the charge on the capacitor is just Q.

Answer 3

Consider the diagram. Let the charge distribution be as shown in the diagram.
The "charge on the capacitor" is said to be the magnitude of the charge on the facing surfaces of the plates. In this case, the "charge on the capacitor" = q.

Now we need to find q. For that we need to know two things:
1. electric field of a single surface of charge
2. electric field inside a metallic plate = 0.

1. electric field of a single surface of charge = \(\sigma/2\epsilon_{0} = Q/(2A\epsilon_{0}) = kQ\), k is taken as a constant for simplicity of calculation.
This can be derived using Gauss Law.

2. Applying the fact that the electric field inside a conductor is zero, we get
\(-k(Q_{1}-q) + kq + k(-q) + k(-Q_{2}+q) = 0\)

\(2kq = k(Q_{1}+Q_{2})\)

\(q = (Q_{1}+Q_{2})/2\)