Three plates A, B and C each of area 0.1 m^2 are separated by 0.885 mm from each other as show

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2. Electric Potential and Capacitance

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Three plates $A, B$ and $C$ each of area $0.1 \ m^2$ are separated by $0.885\ mm$ from each other as shown in the figure. $A$ $10$ $V$ battery is used to charge the system. The energy stored in the system is

A

$1$ $\mu J$

B

$10^{-1}$ $\mu J$

C

$10^{-2}$ $\mu J$

D

$10^{-3}$ $\mu J$

Solution

Here the three plates will make ( $3-1$ ) $= 2$ capacitors of same capacitance $C$ and these two capacitors are connected in parallel.

here, $C=\frac{A \epsilon_{0}}{d}=\frac{0.1 \times 8.854 \times 10^{-12}}{0.885 \times 10^{-3}}=1 n F$

The equivalent capacitance of the system is $C_{e q}=C+C=2 n F$

Energy stored in this system is $U=\frac{1}{2} C_{e q} V^{2}=\frac{1}{2} \times 2 \times 10^{-9} \times 10^{2}=10^{-7} J=10^{-1} \mu J$

Standard 12

Physics

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