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.You need to login to perform this action.
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