An isolated conducting sphere of capacitance 100 pF is charged to a potential of 100 V. A second identical uncharged metal sphere is brought into contact with it. After charge redistribution, the two identical spheres are at the same potential. If the magnitude of the decrease in total electrostatic energy is α × 10⁻⁷ J, what is α?
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Question

An isolated conducting sphere of capacitance 100 pF is charged to a potential of 100 V. A second identical uncharged metal sphere is brought into contact with it. After charge redistribution, the two identical spheres are at the same potential. If the magnitude of the decrease in total electrostatic energy is α × 10⁻⁷ J, what is α?

Options

G1 Physics · Accepted Answer

Initial electrostatic energy is Uᵢ = (1 ÷ 2)CV². Thus, Uᵢ = (1 ÷ 2)(100 × 10⁻¹²)(100)² = 5 × 10⁻⁷ J. The spheres are identical, so after contact the total charge is shared equally and the common potential becomes V(final) = 50 V. Final total energy of the two spheres is U(final) = 2×(1 ÷ 2)C(50)² = 2.5 × 10⁻⁷ J. Therefore the decrease in energy is ΔU = Uᵢ-U(final) = 2.5 × 10⁻⁷ J = (5 ÷ 2) × 10⁻⁷ J. Hence α = 5 ÷ 2. Correct option: B.