question_answer

To charges \[{{q}_{1}}\] and \[{{q}_{2}}\] are placed \[30\,\,cm\] apart, shown in the figure. A third charge \[{{q}_{3}}\]is moved along the arc of a circle of radius \[40\,cm\] from \[C\] to D. The change in the potential energy of the system is \[\frac{{{q}_{3}}}{4\pi {{\varepsilon }_{0}}}k\], where \[k\] is [CBSE PMT 2005]

A)            \[8\,{{q}_{2}}\]

B)            \[8\,{{q}_{1}}\]

C)            \[6{{q}_{2}}\]

D)            \[6{{q}_{1}}\]

Correct Answer: A

Solution :

                     Change in potential energy \[(\Delta U)\text{ }={{U}_{f}}{{U}_{i}}\] Þ  \[\Delta U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \left( \frac{{{q}_{1}}{{q}_{3}}}{0.4}+\frac{{{q}_{2}}{{q}_{3}}}{0.1} \right)-\left( \frac{{{q}_{1}}{{q}_{3}}}{0.4}+\frac{{{q}_{2}}{{q}_{3}}}{0.5} \right) \right]\] Þ \[\Delta U=\frac{1}{4\pi {{\varepsilon }_{0}}}[8{{q}_{2}}{{q}_{3}}]=\frac{{{q}_{3}}}{4\pi {{\varepsilon }_{0}}}(8{{q}_{2}})\] \[\therefore \,\,\,k=\text{ }8{{q}_{2}}\]