Question:
A solid sphere of radius \(R\) carries a charge \(Q + q\) distributed uniformly over its volume. A very small point like piece of it of mass \(m\) gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge \(q\). If it acquires a speed \(v\) when it has fallen through a vertical height \(y\) (see figure), then :
(assume the remaining portion to be spherical).
A solid sphere of radius \(R\) carries a charge \(Q + q\) distributed uniformly over its volume. A very small point like piece of it of mass \(m\) gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge \(q\). If it acquires a speed \(v\) when it has fallen through a vertical height \(y\) (see figure), then :
(assume the remaining portion to be spherical).
(assume the remaining portion to be spherical).
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The remaining sphere has charge \(Q\) and acts as a point charge at its centre. Use energy conservation: gain in KE = loss of gravitational PE + loss of electrostatic PE.
Updated On: Nov 5, 2026
- \(v^2 = y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\)
- \(v^2 = 2y\left[\dfrac{qQR}{4\pi\varepsilon_0 (R+y)^3 m} + g\right]\)
- \(v^2 = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\)
- \(v^2 = y\left[\dfrac{qQ}{4\pi\varepsilon_0 R^2 y m} + g\right]\)
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The Correct Option is C
Solution and Explanation
After the piece (charge \(q\)) leaves, the remaining sphere has charge \(Q\) and behaves as a point charge at the centre for the outside piece.
Initially the piece is at distance \(R\) from the centre, finally at distance \(R + y\).
Electrostatic PE: \(U_i = \dfrac{Qq}{4\pi\varepsilon_0 R}\), \(U_f = \dfrac{Qq}{4\pi\varepsilon_0 (R+y)}\).
Energy conservation: \(\tfrac12 m v^2 = mgy + (U_i - U_f)\).
\(U_i - U_f = \dfrac{Qq}{4\pi\varepsilon_0}\left[\dfrac1R - \dfrac1{R+y}\right] = \dfrac{Qq\,y}{4\pi\varepsilon_0 R(R+y)}\).
\(v^2 = 2gy + \dfrac{2Qq\,y}{4\pi\varepsilon_0 R(R+y)m} = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\).
Answer: (3) \(v^2 = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\)
Initially the piece is at distance \(R\) from the centre, finally at distance \(R + y\).
Electrostatic PE: \(U_i = \dfrac{Qq}{4\pi\varepsilon_0 R}\), \(U_f = \dfrac{Qq}{4\pi\varepsilon_0 (R+y)}\).
Energy conservation: \(\tfrac12 m v^2 = mgy + (U_i - U_f)\).
\(U_i - U_f = \dfrac{Qq}{4\pi\varepsilon_0}\left[\dfrac1R - \dfrac1{R+y}\right] = \dfrac{Qq\,y}{4\pi\varepsilon_0 R(R+y)}\).
\(v^2 = 2gy + \dfrac{2Qq\,y}{4\pi\varepsilon_0 R(R+y)m} = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\).
Answer: (3) \(v^2 = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]\)
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