Question

A parallel plate capacitor has plate area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their final separation is 2d. The work done in this process is –

• A. $\frac { \varepsilon _ { 0 } \mathrm { AV } ^ { 2 } } { 2 \mathrm {~d} }$
• B.
• C.
• D. W = 0

Answer 1

Answer: (A)

$\frac { \varepsilon _ { 0 } \mathrm { AV } ^ { 2 } } { 2 \mathrm {~d} }$

Answer 2

W = ∆U = U_f – U_i = $\frac { \mathrm { Q } ^ { 2 } } { 2 \mathrm { C } _ { \mathrm { f } } } - \frac { \mathrm { Q } ^ { 2 } } { 2 \mathrm { C } _ { \mathrm { i } } } = \frac { \mathrm { Q } ^ { 2 } } { 2 } \left( \frac { 2 \mathrm {~d} } { \varepsilon _ { 0 } \mathrm {~A} } - \frac { \mathrm { d } } { \varepsilon _ { 0 } \mathrm {~A} } \right)$

W =