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 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{q}_{1}}}{OA}+\frac{{{q}_{2}}}{OB}+\frac{{{q}_{3}}}{OC}+\frac{{{q}_{4}}}{OD} \right)\] \[V=9\times {{10}^{9}}\] \[\times \left( \begin{align} & \frac{2\times {{10}^{8}}}{1}+\frac{-2\times {{10}^{-8}}}{1} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{-3\times {{10}^{8}}}{1}+\frac{6\times {{10}^{-8}}}{1} \\ \end{align} \right)\]