question_answer

Four equal charges q are placed at four corners of a square of each side a each. Work done in carrying a charge -q from its centre to infinity is:

A) zero                              

B) \[\frac{\left( 2{{q}^{2}} \right)}{\left( \pi {{\varepsilon }_{0}}a \right)}\]

C) \[\frac{\left( \sqrt{2}\,{{q}^{2}} \right)}{\left( \pi {{\varepsilon }_{0}}a \right)}\]               

D) \[\frac{{{q}^{2}}}{\left( 2\pi {{\varepsilon }_{0}}a \right)}\]

Correct Answer: C

Solution :

(c) \[\frac{\left( \sqrt{2}\,{{q}^{2}} \right)}{\left( \pi {{\varepsilon }_{0}}a \right)}\] Potential at centre \[O={{V}_{0}}=4\times \left( k\,.\,q/r \right)\] Here, r= ((diagonal)/2] \[=\left( a/\sqrt{2} \right)\] \[\therefore \,\,{{V}_{0}}=4\times \left[ q/\left( 4\pi {{\varepsilon }_{0}} \right)/\left( a/\sqrt{2} \right) \right]\] \[=\left[ \left( q/\sqrt{2} \right)/\left( \pi {{\varepsilon }_{0}}a \right) \right]\]             … (1) Work done in shifting (-q) from centre to infinity, \[W=-q\left( {{V}_{\infty }}-{{V}_{0}} \right)=q{{V}_{0}}\] \[\]        [From eq. (1)] \[W=\left[ \left( {{q}^{2}}\sqrt{2} \right)/\left( \pi {{\varepsilon }_{0}}a \right) \right]\]