question_answer

    Four charges\[{{q}_{1}}=2\times {{10}^{-8}}C,{{q}_{2}}=-2\times {{10}^{-8}}C,\]\[{{q}_{3}}=-3\times {{10}^{-8}}C\]and\[{{q}_{4}}=6\times {{10}^{-8}}C\]are placed at four comers of a square of side\[\sqrt{2}m\]. What is the potential at the centre of the square?

A)  270V                                    

B)  300V

C)  Zero                                     

D)  100 V

Correct Answer: A

Solution :

                The potential at a point r is given by \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] \[\therefore \]  \[OA=OB=OC=OD=\frac{AC}{2}=\frac{2}{2}\]                                 \[=1\] Potential at the centre                 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}}{OA}+\frac{{{q}_{2}}}{OB}+\frac{{{q}_{3}}}{OC}+\frac{{{q}_{4}}}{OD} \right)\]                 \[V=9\times {{10}^{9}}\]                 \[\times \left( \begin{align}   & \frac{2\times {{10}^{8}}}{1}+\frac{-2\times {{10}^{-8}}}{1} \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{-3\times {{10}^{8}}}{1}+\frac{6\times {{10}^{-8}}}{1} \\ \end{align} \right)\]