A) \[\frac{kq}{r}\]
B) \[\frac{kq}{{{r}^{2}}}\]
C) \[\frac{kq}{{{r}^{3}}}\]
D) \[\frac{2kq}{{{r}^{3}}}\]
Correct Answer: B
Solution :
If \[E=\frac{kq}{{{r}^{2}}}\] then \[V=-\int_{{}}^{{}}{Edr=\frac{kq}{r}}\] The equi-potential surfaces shown due to the point charge have potential of \[V=\frac{kq}{r}\].You need to login to perform this action.
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