#### Question

Suppose a charge +*Q*_{1} is given to the positive plate and a charge −*Q*_{2} to the negative plate of a capacitor. What is the "charge on the capacitor"?

#### Solution

Given :

Charge on the positive plane = `+Q_1`

Charge on the negative plate = `-Q_2`

To calculate: Charge on the capacitor

Let ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y, respectively.

Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.

Let the area of the plates be A and the permittivity of the free space be `∈_0`.

Now, to determine q in terms of Q_{1} and Q_{2}, we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P. Also, we know that the electric field inside a capacitor is zero.

Electric field due to face I at point P, E1 = `(Q_1 - q)/(2∈_0A)`

Electric field due to face II at point P, E_{2}_{=} `(+q)/(2∈_0A)`

Electric field due to face III at point P, E3 = `(-q)/{2∈_0A)`

Electric field due to face IV at point P, E_{4} = `-((-Q_2+q)/(2∈_0A))` (Negative sign is used as point P lies on the LHS of face IV.)

Since point P lies inside the conductor,

E_{1} + E_{2} + E_{3} + E_{4} = 0

`therefore` `Q_1-q+q-q-(-Q_2+q)` = 0

⇒ q = `(Q_1+Q_2)/2`

Thus , the change on the capacitor is `(Q_1+Q_2)/2`, which is the charge on faces II and III.