What's Your Plate Capacity?

Plates are arranged as shown in the figure such that the effective area between each of them is A A and the effective distance between each pair of neighboring plates is d d . What is the net capacitance between the terminals A and B ?

Clarification : ϵ 0 \epsilon_0 is the permittivity of free space.

ϵ 0 A d \frac{\epsilon _0A}{d} 3 ϵ 0 A d \frac{3\epsilon _0A}{d} 5 ϵ 0 A 2 d \frac{5\epsilon _0A}{2d} 7 ϵ 0 A 6 d \frac{7\epsilon _0A}{6d} 5 ϵ 0 A d \frac{5\epsilon _0A}{d} 6 ϵ 0 A d \frac{6\epsilon _0A}{d} None of these choices

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1 solution

Sparsh Sarode
Jun 5, 2016

The given plates can be taken as a set of capacitors and arranged as shown above

The points 1 , 2 \color{#3D99F6}{1, 2} and 3 , 4 \color{#3D99F6}{3, 4} are at same potential. Hence the the circuit becomes as follows:

The above circuit represents a Wheatstone bridge. Hence the capacitor marked as C \color{#D61F06} C becomes ineffective \color{blueviolet}{ \text{ineffective}} .

\therefore The net capacitance, C e f f C_{eff} is

C e f f = ( C + C ) ( C + C ) = 2 C 2 = C C_{eff}=(C+C)||(C+C)=\dfrac{2C}{2}=C where C = ϵ 0 A d C=\dfrac{\epsilon _0A}{d}

C = ϵ 0 A d \color{#20A900}{\boxed{ \therefore C=\dfrac{\epsilon _0A}{d}}}

Thanks...very clear

Rishi K - 5 years ago

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Welcome :)

Sparsh Sarode - 5 years ago

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