A) \[507\,V\]
B) \[607\,V\]
C) \[550\,V\]
D) \[650\,V\]
Correct Answer: A
Solution :
Net potential (i.e. sum of the potential) at the centre (P) of square is given by \[{{V}_{net}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}}{r}+\frac{{{q}_{2}}}{r}+\frac{{{q}_{3}}}{r}+\frac{{{q}_{4}}}{r} \right]=507\,V\]You need to login to perform this action.
You will be redirected in
3 sec