Ncert Exercises & Solutions

Two charges -q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x(x<<d) perpendicular to the line joining the two fixed charged as shown in figure. Show that q will perform simple harmonic oscillation of time period.

$T = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{1 / 2}$ $T = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{1 / 2}$

Let us elaborate the figure fist.

Given, two charge -q at A and B

$AB = AO + OB = 2 d$

x = small distance perpendicular to O.

i.e., x<d mass of charge Q is. So, force of attraction at P towards A and B are each $F = \frac{q (q)}{4 {πε}_{0} r^{2}},$ where AP=BP=r

Horizontal components of these forces F_n are cancelled out. Vertical components along PO add.

If $∠ APO = O,$ the net force on q along PO is

$F' = 2 F \cos Q$
$= \frac{2 q^{2}}{4 {πε}_{0} r^{2}} (\frac{x}{r})$
$= \frac{2 q^{2} x}{4 {πε}_{0} {(d^{2} + x^{2})}^{3 / 2}}$

When, $x < < d, F' = \frac{2 q^{2} x}{4 {πε}_{0} d^{3}} = Kx$

where, $K = \frac{2 q^{2}}{4 {πε}_{0} d^{3}}$

$\Rightarrow F \propto x$

i.e., force on charge q is proportional to its displacement from the centre O and it is directed towards O.

Hence, motion of charge q would be simple harmonic, where

$ω = \sqrt{\frac{K}{m}}$

and $T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{K}}$

$= 2 π \sqrt{\frac{m 4 {πε}_{0} d^{3}}{2 q^{2}}} = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{1 / 2}$

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