question_answer

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:

A)  the particle executes oscillatory motion

B)  the particle remains stationary

C)  the particle executes, SUM along\[x-\]axis

D)  the particle executes SHM along y-axis

E)  the particle moves on circular path

Correct Answer: C

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\[x<<a,\]then                 \[F=-\frac{2{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{3}}}x\] \[\Rightarrow \]               \[F\propto -x\] Hence, SHM along x-axis.