JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Capacitance

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A sphere of capacitance 100 pF is charged to a potential of 100 V. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is

α \times 1 0^{- 7} J

. The value of

α

is ______. (combined capacitance of spheres is 200 pF).

(A)5
(B) $\frac{5}{2}$ ✓ Correct
(C) $\frac{7}{2}$
(D) $\frac{9}{2}$

Answer: (B)

Solution

The charged sphere stores

U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (100 \times 1 0^{- 12}) (100)^{2} = 5 \times 1 0^{- 7} J

. The charge on it is

Q = C V = 1 0^{- 8} C

. When the identical uncharged sphere touches it, the charge is shared across the combined capacitance

200 pF

, so the common potential is

V^{'} = \frac{Q}{2 C} = 50 V

and the stored energy becomes

U_{f} = \frac{1}{2} (200 \times 1 0^{- 12}) (50)^{2} = 2.5 \times 1 0^{- 7} J

. The energy falls by

Δ U = U_{i} - U_{f} = 2.5 \times 1 0^{- 7} J

, so

α = \frac{5}{2}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Capacitance

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