Question

A charged particle having some mass is resting in equilibrium at a height h above the centre of a uniformly charged non conducting horizontal ring of radius r the force of gravity at downwards the particle will be stable if

Answer 1

The particle only will be stable if H > R/sq.root2 , because in this region the electric field increase if H decreases and decreases of H increases.

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Answer 2

electric field due to non conducting ring along its axis is given by, $E=\frac{kqz}{(z^2+r^2)^{3/2}}$

where z is separation between point of observation to centre of ring, q is charge on ring and r is the radius of ring.

we know, electric field will be maximum only when z = R/√2, means at this point value of force must be maximum. and if we increases the value of z from R/√2 it will decrease and if we decreases the value of z from R/√2 , it will increase.

condition of stable equilibrium,

a small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle be downward.

hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.