JEE Main 2026 Apr 6 Shift 2, Physics Q15: Energy Stored In A Capacitor

JEE Main2026Apr 6, Shift 2Physics

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
$\frac{5}{2}$
C
$\frac{7}{2}$
D
$\frac{9}{2}$

Solution

Initial energy (only sphere 1 charged): $U_{i} = \frac{1}{2} (100 pF) (100)^{2} = 5 \times 1 0^{5}$ pJ.

After contact, charge splits equally so each sphere is at $V /2 = 50$ V. Final energy on each: $\frac{1}{2} (100 pF) (50)^{2}$ ; total over both:

$U_{f} = 2 \times \frac{1}{2} (100) (50)^{2} = 2.5 \times 1 0^{5}$ pJ.

$∣Δ U ∣ = U_{i} - U_{f} = 2.5 \times 1 0^{5}$ pJ $= 2.5 \times 1 0^{- 7}$ J.

So $α = \frac{5}{2}$ .

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JEE Main 2026 Apr 6 Shift 2, Physics Q15: Energy Stored In A Capacitor

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