question_answer

Two equal point charges are fixed at \[x=-a\] and \[x=+a\] on the x-axis. Another point charge Q is placed at the origin. The Change in the electrical potential energy of Q, when it is displaced by a small distance \[x\] along the x-axis, is approximately proportional to [IIT-JEE (Screening) 2002]

A)            x    

B)            \[{{x}^{2}}\]

C)            \[{{x}^{3}}\]                         

D)            \[1/x\]

Correct Answer: B

Solution :

 Initially according to figure (i) potential energy of Q is \[{{U}_{i}}=\frac{2kqQ}{a}\]          ......(i) According to figure (ii) when charge Q is displaced by small distance x then it?s potential energy now \[{{U}_{f}}=kqQ\,\left[ \frac{1}{(a+x)}+\frac{1}{(a-x)} \right]\]\[=\frac{2kqQa}{({{a}^{2}}-{{x}^{2}})}\]  .......(ii) Hence change in potential energy \[\Delta U={{U}_{f}}-{{U}_{i}}=2kqQ\,\left[ \frac{a}{{{a}^{2}}-{{x}^{2}}}-\frac{1}{a} \right]\]\[=\frac{2kqQ{{x}^{2}}}{({{a}^{2}}-{{x}^{2}})}\] Since x << a so \[\Delta U=\frac{2kqQ{{x}^{2}}}{{{a}^{2}}}\Rightarrow \Delta U\propto {{x}^{2}}\]