A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 `mu`C. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 `mu`C. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100V?

Explanation

(i) Intial voltage,
`V_(1)=V` volts,
Charge stored,
`Q_(1)=300 mu C`
`Q_(1)=Cv_(1) " " ` …(i)
Charged potential,
`V_(2)=V-100V.`
`Q_(2)=100 mu C`
`Q_(2)=CV_(2) " " ` ...(ii)
By dividing (ii) from (i), we get
`(Q_(1))/(Q_(2))=(CV_(1))/(CV_(2))rArr (Q_(1))/(Q_(2))=(V_(1))/(V_(2))`
`(300)/(100)=(V)/(V-100)`
`V=150` volts.
`C=(Q_(1))/(V_(1))=(300xx10^(-6))/(150)=2xx10^(-6)F=2 mu F`
(ii) If the voltage applied had increased by 120V, then
`V_(3)=150 +100=250V`
Hence, charge stored in the capacitor,
`Q_(3)=CV_(3)=2xx10^(-6)xx250 =500 mu C`