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 -

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TestHub · Accepted Answer

The system consists of two capacitors in parallel.

Capacitance of one capacitor: $C = \frac{ϵ _{0} A}{d}$

Given $A = 0.1 m^{2}$ , $d = 0.885 \times 1 0^{- 3} m$ , $ϵ_{0} = 8.85 \times 1 0^{- 12} F / m$ .

$C = \frac{8.85 \times 1 0 ^{- 12} \times 0.1}{0.885 \times 1 0 ^{- 3}} = 1 \times 1 0^{- 9} F$ .

Total capacitance $C_{t o t a l} = 2 C = 2 \times 1 0^{- 9} F$ .

Energy stored $U = \frac{1}{2} C_{t o t a l} V^{2} = \frac{1}{2} \times 2 \times 1 0^{- 9} \times (10)^{2} = 1 \times 1 0^{- 7} J = 100 \times 1 0^{- 9} J = 100 n J = 0.1 μ J = 1 0^{- 1} μ J$ .

The final answer is $1 0^{- 1} μ J$ .

Physics - Capacitor Question with Solution | TestHub

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