NEETPhysicsCapacitorMCQ+4 / −1
A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is

- A4.5 106 J
- B3.25 106 J
- C2.25 106 J
- D1.5 106 J
View written solutionFree
Correct answer: C
q1 = CV
= 900 $\times$ 10$-$12 $\times$ 100
= 9 $\times$ 10$-$8 C

$$V = {{{C_1}{V_1} + {C_2}{V_2}} \over {{C_1} + {C_2}}}$$
$$ = {{9 \times {{10}^{ - 8}} + 0} \over {1800 \times {{10}^{ - 12}}}} = {{100} \over 2} = 50$$ V
$$U = {1 \over 2}({C_1} + {C_2}){V^2}$$
$$ = {1 \over 2} \times 1800 \times {10^{ - 12}} \times 50 \times 50$$
$$ = 225 \times {10^{ - 8}}$$
$$U = 2.25 \times {10^{ - 6}}$$ J

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