Three plates of common surface area A are connected as shown. effective capacitance will be

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1. Electric Charges and Fields

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Three plates of common surface area $A$ are connected as shown. The effective capacitance will be

A

$\frac{{{\varepsilon _0}A}}{d}$

B

$\frac{{{3 \varepsilon _0}A}}{d}$

C

$\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}$

D

$\frac{{{2 \varepsilon _0}A}}{d}$

Solution

The given circuit is equivalent to parallel combination of two identical capacitors, each

having capacitance $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}$

Hence $C_{e q}=2 C=\frac{2 \varepsilon_{0} A}{d}$

Standard 12

Physics

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