Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other…
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 figure. Then the capacitance of the system between $a$ and $b$ is
$\frac{3 \varepsilon_{0} A}{d}$
$\frac{2 \varepsilon_{0} A}{d}$
$\frac{2 \varepsilon_{0} A}{3 d}$
$\frac{3 \varepsilon_{0} A}{2 d}$
Solution
The given circuit can be redrawn as \(C_{\text {eq }}=\frac{3 \mathrm{C}}{2}=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}\)
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