Four plates of equal area $A$ are separated by equal distances $d$ and are arranged as shown in the figure. The equivalent capacity is
Medium
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(a) The given circuit is equivalent to a parallel combination two identical capacitors
Hence equivalent capacitance between $A$ and $B$ is
$C$ = $\frac{{{\varepsilon _0}A}}{d} + \frac{{{\varepsilon _0}A}}{d}$
$ = \frac{{2{\varepsilon _0}A}}{d}$
Hence equivalent capacitance between $A$ and $B$ is
$C$ = $\frac{{{\varepsilon _0}A}}{d} + \frac{{{\varepsilon _0}A}}{d}$
$ = \frac{{2{\varepsilon _0}A}}{d}$
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