JEE Main 2020 · 5 September · morning · Single correct · +4 / −1
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) v² = y[ qQ/(4π ε ₀R( R + y )m) + g ]
- (b) v² = 2y[ qQR/(4π ε ₀( R + y )³m) + g ]
- (c) v² = 2y[ qQ/(4π ε ₀R( R + y )m) + g ]
- (d) v² = y[ qQ/(4π ε ₀R²ym) + g ]
Answer and solution
Answer (c) v² = 2y[ qQ/(4π ε ₀R( R + y )m) + g ]
Solution
Physics sir · 4 steps
The electric force weakens as the piece falls, so use energy, not force times distance.
- Remaining sphere. After the piece leaves, the sphere keeps charge Q and stays spherical. By Gauss's law, outside it the field and potential are those of a point charge Q at its centre.
- Start and end. The piece starts at the surface, a distance R from the centre. After falling y, it is at R + y. Both gravity and the repulsion from Q do positive work on the way down.
1/2mv² = mgy + qQ/4πε₀(1/R-1/(R+y)) = mgy + (qQ y)/(4πε₀ R(R+y))
- Solve for v squared. Multiply by 2/m and take y out as a common factor: v² = 2y[qQ/(4πε₀ R(R+y)m) + g].
- So. The correct relation is option (c).
The idea to keep. When a force varies with distance, do not multiply it by the displacement. Use the change in potential energy, and treat any uniformly charged sphere as a point charge at its centre for outside points.
Answer from the NTA final answer key. Solution by PrepWiser.