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A uniformly charged, non-conducting circular disc of radius a lies on the floor with its symmetry axis vertical. A particle of mass m and charge q is released from rest on the axis at a height H above the disc. If it is just able to reach the disc, the electrostatic potential at the starting point is

1. (σ/ϵ0)[(a² + H²)^(1/2) − H]
2. (σ/ϵ0)[(a² + H²)^(1/2) + H]
3. (σ/2ϵ0)[(a² + H²)^(1/2) − H]
4. (σ/2ϵ0)[(a² + H²)^(1/2) + H]

Correct answer: (σ/2ϵ0)[(a² + H²)^(1/2) − H]

Solution

The potential on the axis of a uniformly charged disc at height z is V = (sigma/(2*eps0))*(sqrt(a^2+z^2) - z). At the starting height H, V = (sigma/(2*eps0))*(sqrt(a^2+H^2) - H).

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