R ત્રિજ્યા ધરાવતી રીંગ પર ધન વિદ્યુતભાર Q વિત | vedclass

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Question

When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the rin

Vedclass · Accepted Answer

$\left[\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}\right]^{1 / 2}$

Vedclass · Answer

When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the ring.

$F=q E$ (towards centre)

$=q \cdot \frac{k Q x}{\left(R^{2}+x^{2}\right)^{3 / 2}}$

If $R \gg x,$ then $R^{2}+x^{2} \approx R^{2}$

and $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q x}{R^{3}}$ (towards centre)

$\Rightarrow a=\frac{F}{m}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q x}{m R^{3}}$

since, restoring force $F_{E} \propto x,$ therefore motion of charge particle will be $SHM.$

Time period of $SHM,$

$T=\frac{2 \pi}{\omega}=\left[\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}\right]^{1 / 2}\left[\because \omega^{2}=\frac{Q q}{4 \pi \varepsilon_{0} R^{3}}\right]$

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