NEET 2022 — Physics solution

NEET2022Physics

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:

A
$3.25 \times 1 0^{- 6}$ J
B
$2.25 \times 1 0^{- 6}$ J
C
$1.5 \times 1 0^{- 6}$ J
D
$4.5 \times 1 0^{- 6}$ J

Solution

The charged capacitor initially carries charge $Q = C V = 900 \times 1 0^{- 12} \times 100 = 9 \times 1 0^{- 8}$ C.

When connected to the second uncharged capacitor (same C), charge redistributes until both have the same potential.

Common potential: $V_{C} = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}} = \frac{C \times 100 + C \times 0}{2 C} = 50$ V.

Energy stored after sharing: $U = 2 \times \frac{1}{2} C V_{C}^{2} = C V_{C}^{2} = 900 \times 1 0^{- 12} \times 5 0^{2} = 2.25 \times 1 0^{- 6}$ J.

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