Exams › JEE Advanced › Physics
A solid sphere of radius R carries total charge (Q + q) uniformly distributed through its volume. A tiny point-like fragment of mass m carrying charge q breaks off from the bottom of the sphere and falls vertically under gravity. Treating the remaining sphere as essentially carrying charge Q at radius R, if the fragment attains speed v after falling a vertical distance y, then:
- v² = 2y [ qQ/(4*pi*epsilon0 * R*(R+y)*m) + g ]
- v² = y [ qQ/(4*pi*epsilon0 * R² * y * m) + g ]
- v² = 2y [ qQR/(4*pi*epsilon0 * (R+y)³ * m) + g ]
- v² = y [ qQ/(4*pi*epsilon0 * R*(R+y)*m) + g ]
- v² = 2y [ qQ/(4*pi*epsilon0 * R*(R+y)) + g ]
Correct answer: v² = 2y [ qQ/(4*pi*epsilon0 * R*(R+y)*m) + g ]
Solution
The fragment is repelled by the sphere's charge Q (treated as a point charge at center, distance R initially). Both gravity and the Coulomb force accelerate it; energy conservation gives v².
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