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Q.

A free point charge Q of mass m is kept at the midpoint of two stationary point charges distant 2l and having charge Q each. If the free charge Q is slightly displaced along the line joining stationary charges, then the angular frequency of resulting SHM will be: $(K = \frac{1}{4 {πε}_{0}})$ ,

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a

$\sqrt{\frac{{KQ}^{2}}{{ml}^{3}}}$

b

$\sqrt{\frac{{KQ}^{2}}{2 {ml}^{3}}}$

c

$\sqrt{\frac{{KQ}^{2}}{4 {ml}^{3}}}$

d

$\sqrt{\frac{4 {KQ}^{2}}{{ml}^{3}}}$

answer is D.

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Detailed Solution

$\begin{array}{r} \Rightarrow F = {KQ}^{2} (\frac{1}{(l - x)^{2}} - \frac{1}{(l + x)^{2}}) \\ \Rightarrow F = \frac{{KQ}^{2}}{{(l^{2} - x^{2})}^{2}} ((l + x)^{2} - (l - x)^{2}) \\ \Rightarrow F = \frac{{KQ}^{2}}{{(l^{2} - x^{2})}^{2}} \cdot 4 lx \end{array}$

x<< l; F= $\frac{{KQ}^{2}}{l^{4}} 4 lx$

a = $\frac{4 {KQ}^{2} x}{l^{3} m}$

comparing the above equation with $a = - ω^{2} x$ , we get:

$ω = \sqrt{\frac{4 KQ}{{ml}^{3}}}$

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A free point charge Q of mass m is kept at the midpoint of two stationary point charges distant 2l and having charge Q each. If the free charge Q is slightly displaced along the line joining stationary charges, then the angular frequency of resulting SHM will be: K=14πε0,