In the figure two capacitors and have been shown. Each is charged to a potential but with opposite polarity, so that points a and c are on the side of the respective positive plates of and , and points b and d are on the side of the respective negative plates. Switches and are now closed. We have to find (a) the potential difference between points e and f; (b) the charge on and (c) the charge on .

Solution:             Click For PDF Version

(a)

In the circuit shown in the figure the capacitors

and each have been charged to potential and the switches and are open.

Therefore, the magnitude of the charge on will be

,

and the magnitude of the charge on will be

After the switches and are closed an amount of charge will flow from to till both the capacitors have the same potential across their plates. That is the potential difference between points e and f will be the potential difference across each of the two capacitors.

This implies that

Solving this equation, we find

.

We calculate the potential difference across the capacitor . It will also be equal to the potential difference between e and f after the two switches have been closed. We have

(b)

Charge on will be

,

(c)

And charge on will be