The negative plate of a parallel plate capacitor is given a charge of `-20 X 10^(-8) C`. Find the charges appearing on the four surface of the capacitor plates.

Explanation

Let the charge appearing on the inner surface of the nagative plate be -Q. The charge on its outer surface will be `
` Q - 20 X 10 ^(-8) C`
`. The charge on the inner surface of the positive plate will be + Q from Gauss's law and that on the outer surface will be -Q as the positive plate is electrically neutral. The distribution is shown in figure. To obtain the value of Q, consider the electric field at a point P inside the upper plate. Field due to suface (1) `
`= (Q)/(2 epsilon_0 A)`
` upward, due to surface (2) `
`=(Q)/(2epsilon_0 A)`
` upward, due ot surface (3) `
`= (Q)/(2 epsilon_0 A)`
` downward and due to surface (4) `
`= (Q - 20 X10^(-C)/(2epsilon_0 A)`
` upward. As P is a point inside the conductor, the field here must be zero. Thus, `
` Q = - Q +20 X10^(-8)`
` C. or Q = 10 X 10^(-8)`
`C. The charges on the four surfaces may be written immediately from figure.