question_answer

Between two infinitely long wires having linear charge densities\[\lambda \], and\[-\lambda \], there are two points A and B as shown in figure. The amount of work done by the electric field in moving a point charge \[{{q}_{0}}\] from A to B is equal to:

A)  \[\frac{\lambda {{q}_{0}}}{4\pi {{\varepsilon }_{0}}}In\,2\]                         

B)  \[-\frac{2\lambda {{q}_{0}}}{\pi {{\varepsilon }_{0}}}In\,2\]

C)  \[\frac{2\lambda {{q}_{0}}}{\pi {{\varepsilon }_{0}}}In\,2\]                         

D)  \[\frac{\lambda {{q}_{0}}}{\pi {{\varepsilon }_{0}}}In\,2\]

Correct Answer: D

Solution :

            The electric field at point P is:             \[E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}x}+\frac{\lambda }{2\pi {{\varepsilon }_{0}}(3a-x)}\]             \[=\,\frac{\lambda }{2\pi {{\varepsilon }_{0}}}\left[ \frac{1}{x}-\frac{1}{3a-x} \right]\] The work done in moving chare \[{{q}_{0}}\] from A to B is: \[\int_{a}^{2a}{{{q}_{0}}Edx=\frac{{{q}_{0}}\lambda }{2\pi {{\varepsilon }_{0}}}\int_{a}^{2a}{\left( \frac{1}{x}-\,\frac{1}{3a-x} \right)\,dx}}\] \[=\,\frac{{{q}_{0}}\lambda }{2\pi {{\varepsilon }_{0}}}\,\left[ \ln \,x-\ln \,(3a-x) \right]_{a}^{2a}\] \[=\frac{{{q}_{0}}\lambda }{2\pi {{\varepsilon }_{0}}}\left[ \ln \,\frac{x}{3a-x} \right]_{a}^{2a}\] \[=\frac{{{q}_{0}}\lambda }{2\pi {{\varepsilon }_{0}}}\,\left[ \ln 2-\ln \,\frac{1}{2} \right]=\frac{\lambda {{q}_{0}}}{\pi {{\varepsilon }_{0}}}\,\ln 2\]