Four metallic plates each of surface area (of one side) $A$, are placed at a distance $d$ apart from each other. The two outer plates are connected to a point $P$ and the two inner plates are connected to another point as shown in figure below. Then, the capacitance of the system is
KVPY 2009, Medium
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(c)
Above system is equivalent to two capacitors in parallel,
So, capacitance of system is
$C_{ eq }=C_1+C_2=\frac{\varepsilon_0 A}{d}+\frac{\varepsilon_0 A}{d}=\frac{2 \varepsilon_0 A}{d}$
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