A charged particle having some mass is restin | vedclass

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Question

$F=\frac{k Q x(+q)}{\left(R^{2}+x^{2}\right)^{3 / 2}}-m g$

if $\frac{\mathrm{dF}}{\mathrm{dx}}<0$

Then the particle is in stable equilibrium

Vedclass · Accepted Answer

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

Vedclass · Answer

$F=\frac{k Q x(+q)}{\left(R^{2}+x^{2} ight)^{3 / 2}}-m g$ if $\frac{\mathrm{dF}}{\mathrm{dx}}<0$ Then the particle is in stable equilibrium $\Rightarrow \frac{\left(R^{2}+x^{2} ight)^{3 / 2}-\frac{3}{2}\left(R^{2}+x^{2} ight)^{1 / 2}\left(2 x^{2} ight)}{\left(R^{2}+x^{2} ight)^{3}}<0$ $\Rightarrow \mathrm{R}^{2}+\mathrm{x}^{2}-3 \mathrm{x}^{2}<0$ $\Rightarrow x>\frac{R}{\sqrt{2}}$ or $x<-\frac{R}{\sqrt{2}}$ $ herefore$ Only if $x>\frac{R}{\sqrt{2}}$ The equilibrium will be stable.

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 $R$

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