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  • 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

Two charges –q each are fixed and 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 charges as shown in Fig. 1.14. Show that q will perform simple harmonic oscillation of time period.



 

Answers (1)

Explanation:-

Let the charge q be displaced slightly by x(x<<d)perpendicular to the line joining the two fixed charges. The net force on the charge q will be towards O.The motion of charge -q to be simple harmonic if the force on charge q must be proportional to its distance from the center O and is directed towards O.
The net force on the charge F_{net}= 2F \cos\theta
Here  F=\frac{1}{4\pi \varepsilon _0}\frac{q(q)}{r^2}= \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )}

And\cos \theta =\frac{x}{\sqrt{x^2+d^2}}

Hence, F_{net}=2 \left [ \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )} \right ]\left [ \frac{x}{\sqrt{x^2+d^2}} \right ]

= \frac{1}{2\pi \varepsilon _0}\frac{q^2x}{\left ( d^2+x^2 \right )^{3/2}}=\frac{1}{2\pi \varepsilon _0}\frac{q^2x}{d^3\left ( 1+\frac{x^2}{d^2} \right )^{3/2}}

As x<<d, then F_{net}=\frac{1}{2 \pi \varepsilon _0}\frac{q^2x}{d^3}\: or \: F_{net}=Kx
i.e. force on charge q is proportional to its displacement from the centre O and it is directed towards O.

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

\omega = \sqrt{\frac{K}{m}}\\ and \: T=\frac{2 \pi }{\omega}=2 \pi\sqrt{\frac{m}{K}}

so, the time period 
\Rightarrow T= 2 \pi \sqrt{\frac{m.4 \pi \varepsilon _0d^3}{2q^2}}=\left [ \frac{8 \pi^3 \varepsilon _0md^3}{q^2} \right ]^{1/2}    

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