Four metallic plates, each with a surface area of one side A , are placed at a distance d from

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1. Electric Charges and Fields

normal

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. The capacitance of the system between $a$ and $b$ is

$\frac{{3{\varepsilon _0}A}}{d}$

$\frac{{2{\varepsilon _0}A}}{d}$

$\frac{2}{3}\frac{{{\varepsilon _0}A}}{d}$

$\frac{3}{2}\frac{{{\varepsilon _0}A}}{d}$

The equivalent arrangement is shown in figure. Capacitance between $\mathrm{A}$ and $\mathrm{B}$ is given by :

$\frac{1}{C_{A B}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}$

or $\quad \mathrm{C}_{\mathrm{AB}}=\frac{\mathrm{C}}{2}$

$\therefore \mathrm{C}_{\mathrm{ab}}=\mathrm{C}+\frac{\mathrm{C}}{2}=\frac{3}{2} \mathrm{C}$

$ = \frac{3}{2} \times \frac{{{ \in _0}A}}{d}$

Standard 12

Physics

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normal

normal

normal

normal