ElectrostaticsMedium
Question
Five positive charges each having charge $q$ are placed at the vertices of a pentagon as shown in the figure. The electric potential ( $V$ ) and the electric field ( $\overrightarrow{E}$ ) at the center O of the pentagon due to these five positive charges are :
Options
A.$V = \frac{5q}{4\pi\varepsilon_{0}r}$ and $\overrightarrow{E} = 0$
B.$V = \frac{5q}{4\pi\varepsilon_{0}r}$ and $\overrightarrow{E} = \frac{5\sqrt{3}q}{8\pi\varepsilon_{0}r^{2}}\widehat{r}$
C.$V = \frac{5q}{4\pi\varepsilon_{0}r}$ and $\overrightarrow{E} = \frac{5q}{4\pi\varepsilon_{0}r^{2}}\widehat{r}$
D.$V = 0$ and $\overrightarrow{E} = 0$
Solution
Electric potential $\rightarrow V = \frac{5kq}{R}$
As regular polygon $\rightarrow \overrightarrow{E} = 0$
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