NEET 2026, Physics Q13: Energy Stored In A Capacitor

NEET2026Physics

Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :

A
0.5 J
B
$1.0 \times 1 0^{- 6}$ J
C
$0.5 \times 1 0^{- 6}$ J
D
1.0 J

Solution

Energy lost when two capacitors are connected is

$Δ H = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$

$= \frac{1}{2} \times \frac{200 \times 200}{200 + 200} \times (100 - 0)^{2} \times 1 0^{- 12}$

$= 50 \times 1 0^{4} \times 1 0^{- 12}$ J $= 5 \times 1 0^{- 7}$ J $= 0.5 \times 1 0^{- 6}$ J.

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NEET 2026, Physics Q13: Energy Stored In A Capacitor

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