question_answer

Charges of \[+\frac{10}{3}\times {{10}^{-9}}C\] are placed at each of the four corners of a square of side \[8\,cm\]. The potential at the intersection of the diagonals is                                                                                         [BIT 1993]

A)                    \[150\sqrt{2}\,volt\] 

B)            \[1500\sqrt{2}\,volt\]

C)                    \[900\sqrt{2}\,volt\] 

D)            \[900\,volt\]

Correct Answer: B

Solution :

             Potential at the centre O,       \[V=4\times \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{a/\sqrt{2}}\] where \[Q=\frac{10}{3}\times {{10}^{-9}}C\]  and \[a=8\,cm\,=\,\,8\times {{10}^{-2}}m\] So \[V=5\times 9\times {{10}^{9}}\times \frac{\frac{10}{3}\times {{10}^{-9}}}{\frac{8\times {{10}^{-2}}}{\sqrt{2}}}\]\[=1500\sqrt{2}\,volt\]