Calculate the capacity of unknown capacitance is connected acrosss a battery of V volts. The charge stored in it is `360 muC`. When potential across the capacitor is reduced by 120V, the charge stored in it becomes `120 muC`.
Calculate (i) the potential V and unknown capacitance C. (ii) What will be the charge stored in the capacitor. If the voltage applied had increased by 120 V

Explanation

Initial voltage, `V_(1) = V ` volts and charge stored,
`Q_(1) = 360 mu C`
`Q_(1) = CV_(1). " " ` …(i)
Changed potential, `V_(2)= V- 120`
`Q_(2) = 120 mu C`
`Q_(2) = CV_(2) " " `...(ii)
By dividing (ii) from (i), we get,
`(Q_(1))/(Q_(2))=(CV_(1))/(CV_(2))rArr (360)/(12) = (V)/(V-120)`
`V= 180` volts.
`C= (Q_(1))/(V_(1))=(360xx10^(-6))/(180)=2 xx 10^(-6) F=2 mu F.`
(ii) If the voltage applied had increased had increased by 120 V, then `V_(3) = 180+120=300V. `
Hence, charge stored in the capacitor ,
`Q_(3) = CV_(3) = 2xx 10^(-6) xx 300 = 600 mu C. `