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)] \[W=\left[ \left( {{q}^{2}}\sqrt{2} \right)/\left( \pi {{\varepsilon }_{0}}a \right) \right]\]You need to login to perform this action.
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