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 \[ABC\] = \[\pi /2\] and \[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\]

A) \[9J\]    

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

C) \[\frac{9}{25}J\]           

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

Correct Answer: D

Solution :

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