Electrostatics Part 3

Concept:Electric potential is a scalar, so it adds algebraically.Electric field is a vector, so it adds according to direction.Explanation:The distance from each vertex to centre $O$ is same, i.e., $r$.Potential due to one charge at $O$: $V_i = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.Total potential is algebraic sum of five equal potentials: $V = 5 \cdot \frac{1}{4 \pi \epsilon_0} \frac{q}{r} = \frac{5q}{4 \pi \epsilon_0 r}$.For electric field, each charge produces field at $O$ of magnitude $E_i = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$.These five vectors point along lines from $O$ to the vertices.Due to symmetry of a regular pentagon, these five equal vectors sum to zero (closed polygon rule).Hence, net electric field $\vec{E} = 0$.Answer:Option B: $V = \frac{5q}{4 \pi \epsilon_0 r}$ and $\vec{E} = 0$.