A positive charge Q is distributed uniformly | vedclass

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Question

(A) The electric field $E$ on the axis of a ring of radius $R$ at a distance $x$ from the centre is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The electric field $E$ on the axis of a ring of radius $R$ at a distance $x$ from the centre is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q x}{(R^{2} + x^{2})^{3/2}}$.The force on the particle of charge $-q$ is $F = -qE = -\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q x}{(R^{2} + x^{2})^{3/2}}$.Given $x \ll R$,we can approximate $R^{2} + x^{2} \approx R^{2}$. Thus,$F \approx -\left( \frac{Q q}{4 \pi \varepsilon_{0} R^{3}} \right) x$.This is the equation of simple harmonic motion $(F = -kx)$,where the effective spring constant $k = \frac{Q q}{4 \pi \varepsilon_{0} R^{3}}$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{Q q}{4 \pi \varepsilon_{0} R^{3} m}}$.The time period $T$ is $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{4 \pi \varepsilon_{0} R^{3} m}{Q q}} = \sqrt{\frac{4 \pi^{2} \cdot 4 \pi \varepsilon_{0} R^{3} m}{Q q}} = \sqrt{\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}}$.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

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