A capacitor of unknown capacitance is connected across a battery of V volts . The charge stored in it is `360 mu C` . When potential across the capacitor is reduced by 120 V , the charge stored in it becomes 120 `mu C ` . Calculate :
The potential V and the unknown capacitance C .

Explanation

Let the capacitance of given capacitor be C mF . Then as per data given in the question :
CV = `360 mu C " " …. (1)` and `C ( V - 120) = 120 mu C " " …. (2)`
Substracting (2) from (1) , we have
`C xx 120 = 240 mu C implies C = (240 mu C)/(120 V) = 2 mu F`
and from (1) `V = (360 mu C)/(C) = (360 mu C)/(2 mu F) = 180 V `