Question

A parallel plate capacitor C₀ is charged to a potential V₀. The energy stored in the capacitor when the charging battery is kept connected and the plate separation is doubled is E₁. The energy stored in the capacitor when the charging battery is disconnected and the separation between plates is doubled is E₂, then E₁/E₂ is-

• A. 4
• B. ¼
• C. 2
• D. ½

Answer 1

Answer: (A)

4

Answer 2

Q₀ = C₀V₀ = Initial charge on capacitor

Battery connected, V ® unchanged

As separation is doubled C = C₀/2

E₁ = $\frac { 1 } { 2 } \left( \frac { \mathrm { C } _ { 0 } } { 2 } \right) \mathrm { V } _ { 0 } ^ { 2 } = \frac { \mathrm { C } _ { 0 } \mathrm {~V} _ { 0 } ^ { 2 } } { 4 }$

Battery disconnected, Charge ® unchanged

As separation is doubled C = C₀/2

E₂ = $\frac { \mathrm { Q } _ { 0 } ^ { 2 } } { 2 \mathrm { C } } = \frac { \mathrm { C } _ { 0 } ^ { 2 } \mathrm {~V} _ { 0 } ^ { 2 } } { 2 \cdot \frac { \mathrm { C } _ { 0 } } { 2 } }$ = C₀V₀²

\ $\frac { E _ { 1 } } { E _ { 2 } } = \frac { 1 } { 4 }$