question_answer

Two point charges \[100\,\mu \,C\] and \[5\,\mu \,C\] are placed at points \[A\] and \[B\] respectively with \[AB=40\,cm\]. The work done by external force in displacing the charge \[5\,\mu \,C\] from \[B\] to \[C\], where \[BC=30\,cm\], angle \[ABC=\frac{\pi }{2}\] and \[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\]                                                               [MP PMT 1997]

A)                    \[9\,J\]

B)                                      \[\frac{81}{20}J\]

C)                    \[\frac{9}{25}J\]

D)                                      \[-\frac{9}{4}J\]

Correct Answer: D

Solution :

                     Work done in displacing charge of 5 m C from B to C is \[W=5\times {{10}^{-6}}\,({{V}_{C}}-{{V}_{B}})\] where \[{{V}_{B}}=9\times {{10}^{9}}\times \frac{100\times {{10}^{-6}}}{0.4}=\frac{9}{4}\times {{10}^{6}}V\] and \[{{V}_{C}}=9\times {{10}^{9}}\times \frac{100\times {{10}^{-6}}}{0.5}=\frac{9}{5}\times {{10}^{6}}V\] So \[W=5\times {{10}^{-6}}\times \left( \frac{9}{5}\times {{10}^{6}}-\frac{9}{4}\times {{10}^{6}} \right)=-\frac{9}{4}J\]