Question

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C of mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it? (b) Assuming x<<d, show that this force is proportional to x. (c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.

Answer 1

(a) The charge q is displaced by a distance x on the perpendicular bisector of AB.
As shown in the figure, the horizontal component of the force is balanced.

$\sin \theta =\frac{x}{\sqrt{{\left({\displaystyle \frac{d}{2}}\right)}^2+x^2}}$
Total vertical component of the force, $F\text{'}=2F\sin \theta$
$$\begin{gathered} F\text{'}=2\times \frac{1}{4\mathrm{\pi }{\epsilon }_0}\times \frac{qQ}{{\left({\displaystyle \frac{d}{2}}\right)}^2+x^2}\times \frac{x}{\sqrt{{\left({\displaystyle \frac{d}{2}}\right)}^2+x^2}} \\ \Rightarrow F\text{'}=\frac{1}{2\mathrm{\pi }{\epsilon }_0}\times \frac{qQx}{{\left[{\left({\displaystyle \frac{d}{2}}\right)}^2+x^2\right]}^{3/2}} \end{gathered}$$
This is the net electric force experienced by the charge q.

(b) When x < < d:
$$\begin{gathered} F\text{'}=\frac{1}{2{\mathrm{\pi \epsilon }}_0}\frac{qQx}{{\left({\displaystyle \frac{d}{2}}\right)}^3}\left(\because x^2<<{\left(\frac{d}{2}\right)}^2\right) \\ \Rightarrow F\text{'}=\frac{4}{{\mathrm{\pi \epsilon }}_0}\frac{qQx}{d^3} \end{gathered}$$

(c) For the particle to execute simple harmonic motion:
F' = mw²x
$$\begin{gathered} \Rightarrow \frac{4}{\pi {\epsilon }_0}\frac{qQx}{d^3}=m{\left(\frac{2\pi }{T}\right)}^{2}x \\ \Rightarrow T^2=\frac{m{\pi }^3{\epsilon }_0{d}^3}{Qq} \\ \Rightarrow T={\left[{\displaystyle \frac{m{\pi }^3{\epsilon }_0{d}^3}{Qq}}\right]}^{1/2} \end{gathered}$$