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Question: The capacitance of the system of parallel plate capacitor shown in the figure is <img src="https://...

The capacitance of the system of parallel plate capacitor shown in the figure is

A

2ε0 A1 A2( A1+A2)d\frac { 2 \varepsilon _ { 0 } \mathrm {~A} _ { 1 } \mathrm {~A} _ { 2 } } { \left( \mathrm {~A} _ { 1 } + \mathrm { A } _ { 2 } \right) \mathrm { d } }

B

C

ε0 A1 d\frac { \varepsilon _ { 0 } \mathrm {~A} _ { 1 } } { \mathrm {~d} }

D

ε0 A2 d\frac { \varepsilon _ { 0 } \mathrm {~A} _ { 2 } } { \mathrm {~d} }

Answer

ε0 A1 d\frac { \varepsilon _ { 0 } \mathrm {~A} _ { 1 } } { \mathrm {~d} }

Explanation

Solution

Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the fringe effect, the effective area of the plate of area A2 is A1. Thus the capacitance between the plates is C=ε0 A1 d\mathrm { C } = \frac { \varepsilon _ { 0 } \mathrm {~A} _ { 1 } } { \mathrm {~d} }