Each of the three plates shown in figure has an area of 200 cm^2 on one side and the gap between the adjacent plates is 0.2 mm. The emf of the battery is 20V. Find the distribution of charge on various surfaces of the plates. What is the equivalent capacitance of the system between the terminal points?

Explanation

Suppose the nagative terminal of the battery gives a charge - Q to the plate B. As the situation is symmetric on the two sides of B, the two faces of the plates B will share equal charges `
`-Q/2`
`each. As the positive terminal of the battery has supplied just this much charge (+ Q) to A and C, the outer surfaces of A and C will have no charge. The distribution will be as shown in figure. The capacitance between the plates A and B is `
`8.85 X 10^(-12) Fm(-1) X (200 X 10^(-4) m^2)/(2X10^(-4) m) = 8.85 X 10 ^(10) F = 0.885 nF. Thus, (Q)/(2) = 0.885 mF X 20 V = 17.7 nC. The distribution of charge on various surfaces may be the equivalent capacitance is `
` (Q)/(20 V) = 1.77 nF.