Four plates of the same area of cross-section are joined as shown in the figure. The distance between each plate is $d$. The equivalent capacity across $A$ and $B$ will be
A$\frac{{2{\varepsilon _0}A}}{d}$
B$\frac{{3{\varepsilon _0}A}}{d}$
C$\frac{{3{\varepsilon _0}A}}{{2d}}$
D$\frac{{{\varepsilon _0}A}}{d}$
Medium
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B$\frac{{3{\varepsilon _0}A}}{d}$
b (b) The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance $ = 3C = 3\frac{{{\varepsilon _0}A}}{d}$
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