Given:

Charge on positive plate=Q₁

Charge on negative plate=Q₂

Assume a rectangular gaussian surface ABCD as shown in fig.

Let the charge on the capacitor plates be “q” and the area of plates be A. Then,

Charge appearing on face 1=Q₁-q.

Charge appearing on face 2=q.

Similarly,

Charge appearing on face 3= -q.

Since, the total charge enclosed by a closed surface =0)

Charge appearing on face 4=Q₂ +q.

Formula used:

We know that,

I) Electric field inside any conductor=0.

∴ Electric field at point Pinside plate)=0.

E₁+E₂+E₃+E₄=0 …i)

This Electric field is the net effect of fields at point P due to faces I, II, III and IV.

II) Electric field due a thin sheet, E=

Where

E is the electric filed due to thin plate

Q is the total charge enclosed in the gaussian surface

A is the area of the plate

ε_o is the permittivity of the vacuum

Thus, Electric field at point P due to face I E₁=

Electric field at point P due to face II E₂=

Electric field at point P due to face III E₃=

Electric field at point P due to face IV E₄=

negative sign because electric field due to face IV is in leftwards direction).

Putting the values in equation (i) we get,

On solving the above equation, we get

Q₁-q+q-q-Q₂-q=0

Q₁-Q₂-2q=0

q

Thus, the charge on the capacitor is