A parallel plate capacitor of capacity C0 is charged to a potential V0. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates doubled is E2. The ratio E1 / E2 is.

Question

A parallel plate capacitor of capacity C0 is charged to a potential V0. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates doubled is E2. The ratio E1 / E2 is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Initial energy: E=(1/2) C0 V02 (i) E1=2 E, because charge remains same and capacitance becomes half. (ii) E2=E / 2, because potential remains same and capacitance becomes half. ⇒ (E1/E2)=4

Initial energy: E=21​C0​V02​ (i) E1​=2E, because charge remains same and capacitance becomes half. (ii) E2​=E/2, because potential remains same and capacitance becomes half. ⇒E2​E1​​=4

Initial energy: $E = \frac{1}{2} C_{0} V_{0}^{2}$
(i) $E_{1} = 2 E$ , because charge remains same and capacitance becomes half.
(ii) $E_{2} = E /2$ , because potential remains same and capacitance becomes half.
$\Rightarrow \frac{E _{1}}{E _{2}} = 4$