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(B) The electric field on the axis of a uniformly charged ring at a distance $x$ is $E = \frac{k Q x}{(R^2 + x^2)^{3/2}}$.For a particle of charge $q$

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only if $H > \frac{R}{\sqrt{2}}$

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(B) The electric field on the axis of a uniformly charged ring at a distance $x$ is $E = \frac{k Q x}{(R^2 + x^2)^{3/2}}$.For a particle of charge $q$ and mass $m$ in equilibrium,the net force $F = qE - mg = 0$.For stable equilibrium,the condition is $\frac{dF}{dx} Calculating the derivative: $\frac{dF}{dx} = q \frac{dE}{dx} = qkQ \left[ \frac{(R^2 + x^2)^{3/2} - x \cdot \frac{3}{2}(R^2 + x^2)^{1/2} \cdot 2x}{(R^2 + x^2)^3} \right] Simplifying the expression: $(R^2 + x^2) - 3x^2 $R^2 - 2x^2  R^2 \Rightarrow x > \frac{R}{\sqrt{2}}$.Since the particle is at height $H$,the equilibrium is stable if $H > \frac{R}{\sqrt{2}}$.

$A$ 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 acts downwards. The equilibrium of the particle will be stable if:

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