A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?

Sol. Given that C = 100 PF = 100 * 10^-12 F C base cq = 20 PF = 20 * 10^-12 F V = 24 V q = 24 * 100 * 10^-12 = 24 * 10^-10 Q base 2 = ? Let q base 1 = The new charge 100 PF V base 1 = The Voltage. After the flow of charge, potential is same in the two capacitor V base 1 = q base 2/C base 2 = q base 1/C base 1 = q – q base 1/C base 2 = q base 1/C base 1 = 24 * 10^-10 –q base 1/24 * 10^-12 = q base 1/100 * 10^-12 = 24 * 10^-10 – q base 1 = q base 1/5 = 6q base 1 = 120 * 10^-10 = q base 1 = 120/6 * 10^-10 = 20 * 10^10 ∴ V base 1 = q base 1/ C base 1 = 20 *10^-10/100 * 10^-12 = 20 V