Q. Two equal charges q are kept fixed at a and +a along the $ x- $ axis. A particle of mass m and change $ \frac{q}{2} $ is brought to the origin and given a small displacement along the $ x- $ axis, then:
KEAMKEAM 2000
Solution:
From Coulombs law
$ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $ $ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q\times q/2}{{{(a+x)}^{2}}}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q\times q/2}{{{(a-x)}^{2}}} $ $ =\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2}\left[ \frac{1}{{{(a+x)}^{2}}}-\frac{1}{{{(a-x)}^{2}}} \right] $ $ =\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2}\left[ \frac{4ax}{{{({{a}^{2}}-{{x}^{2}})}^{2}}} \right] $ When $ x<
