question_answer

Two equal -ve charge -q are fixed at the point (0, a) and (0,-a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the\[x-\]axis. The charge will:

A)  execute SHM about the origin

B)  move to the origin and remain at rest

C)  c) move to infinity

D)  execute oscillatory but not SHM

Correct Answer: D

Solution :

                 Key Idea: Components offerees on charge Q due to\[-q\] and\[+q\]will cancel along y-axis and along\[x-\]axis they will add up. By symmetry of problem the components of forces on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards 0. If at any time charge Q is at a distance\[x\]from 0, \[F\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-qQ}{({{a}^{2}}+{{x}^{2}})}\]                                                 \[\times \frac{x}{{{({{a}^{2}}+{{x}^{2}})}^{1/2}}}\] i.e.,        \[F=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQx}{{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}\] Since, the restoring force F is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic. Note: If the point C is close to 0 so. that\[x<<a,\] \[F=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{{{a}^{3}}}x=-kx\]with\[k=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{2qQ}{{{a}^{3}}}\] i.e., the restoring force will become linear and so the motion is simple harmonic with time period \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{4\pi {{\varepsilon }_{0}}m{{a}^{3}}}{2qQ}}\]