Four metallic plates each with a surface area of one side $A$ are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is
A$\frac{{3{\varepsilon _0}A}}{d}$
B$\frac{{2{\varepsilon _0}A}}{d}$
C$\frac{{2{\varepsilon _0}A}}{{3d}}$
D$\frac{{3{\varepsilon _0}A}}{{2d}}$
Medium
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D$\frac{{3{\varepsilon _0}A}}{{2d}}$
d (d) The given circuit can be redrawn as follows
$==>$ ${C_{eq}} = \frac{{3C}}{2} = \frac{{3{\varepsilon _0}A}}{{2d}}$
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