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JEE Main 2026 · 6 Apr · Shift 2, Physics question 40

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×10−7\alpha\times10^{-7} J. The value of α\alpha is ________.

(combined capacitance of spheres is 200 pF)

Choose one option
Answer and worked solution

Official answer: B

Before. U1=12(100 pF)(100)2=5×10−7U_1=\tfrac12(100\text{ pF})(100)^2=5\times10^{-7} J, with Q=10−8Q=10^{-8} C.

After. The charge spreads over 200 pF:
U2=Q22C=10−164×10−10=2.5×10−7 J.U_2=\frac{Q^2}{2C}=\frac{10^{-16}}{4\times10^{-10}}=2.5\times10^{-7}\text{ J}.

Change: 2.5×10−72.5\times10^{-7} J, so α=52\alpha=\tfrac52.

(B)

Topics:Energy stored in a capacitor
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