Question

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:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?

Answer 1

Given - initial voltage $V_1=V$ , initial charge $q_1=360\mu C$ ,

therefore , capacitance $C=q_1/V_1=360/V$ , ................eq1

when , voltage $V_2=V-120$ , charge $q_2=120\mu C$ ,

therefore , capacitance $C=q_2/V_2=120/(V-120)$ , ..................eq2

capacitance of the capacitor remains constant as there is no change in area of the plates , medium and distance between plates ,

(i) Dividing eq1 by eq2 ,

$360/V=120/(V-120)$ ,

or $3(V-120)=V$ ,

or $V=360/2=180volt$

(ii) Capacitance $C=360/180=2\mu F$

now the new voltage is $V^{\prime }=180+120=300V$ ,

therefore charge stored $q^{\prime }=C{V}^{\prime }=2\times 300=600\mu C$