Ncert Exercises & Solutions

Total charge -Q is uniformly spread along length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.

(a) Show that the particle executes a simple harmonic oscillation.

(b) Obtain its time period.

Let us draw the figure according to question,

A gentle push on q along the axis of the ring gives rise to the situation shown in the figure below.

Taking line elements of charge at A and B, having unit length, then charge on each elements.

$dF = 2 (- \frac{Q}{2 πR}) q \times \frac{1}{4 {πε}_{0}} \frac{1}{r^{2}} \cos θ$ $dF = 2 (- \frac{Q}{2 πR}) q \times \frac{1}{4 {πε}_{0}} \frac{1}{r^{2}} \cos θ$

Total force on the charge q, due to entire ring

$F = - \frac{Qq}{πR} (πR) . \frac{1}{4 {πε}_{0}} \frac{1}{r^{2}} . \frac{2}{r}$ $F = - \frac{Qq}{πR} (πR) . \frac{1}{4 {πε}_{0}} \frac{1}{r^{2}} . \frac{2}{r}$
$F = - \frac{Qqz}{4 {πε}_{0} {(Z^{2} + R^{2})}^{3 / 2}}$ $F = - \frac{Qqz}{4 {πε}_{0} {(Z^{2} + R^{2})}^{3 / 2}}$

Here, Z<<R,

$F = - \frac{Qqz}{4 {πε}_{0} R^{3}} = - Kz$ $F = - \frac{Qqz}{4 {πε}_{0} R^{3}} = - Kz$

where

$\frac{Qq}{4 {πε}_{0} R^{3}} = constant$ $\frac{Qq}{4 {πε}_{0} R^{3}} = constant$

$\Rightarrow F \propto - Z$ $\Rightarrow F \propto - Z$

Clearly, force on q is proportional to negative of its displacement. Therefore, motion of q is simple harmonic.

$ω = \sqrt{\frac{K}{m}} and T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{K}}$ $ω = \sqrt{\frac{K}{m}} and T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{K}}$
$T = 2 π \sqrt{\frac{m 4 {πε}_{0} R^{3}}{Qq}}$ $T = 2 π \sqrt{\frac{m 4 {πε}_{0} R^{3}}{Qq}}$
$\Rightarrow T = 2 π \sqrt{\frac{4 {πε}_{0} {mR}^{3}}{Qq}}$ $\Rightarrow T = 2 π \sqrt{\frac{4 {πε}_{0} {mR}^{3}}{Qq}}$

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