Electrostatics Part 3

Concept:Electric potential is a scalar quantity, so the potentials due to all the charges add algebraically.Electric field is a vector quantity, so the individual fields must be added as vectors, keeping their directions in mind.Potential at the centre $O$:In the regular pentagon shown, every vertex is at the same distance $r$ from the centre $O$.Potential due to one charge $q$ at the centre is $V_i=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$.Since $V$ is a scalar, the five equal contributions simply add:$V=5\times\frac{1}{4\pi\epsilon_0}\frac{q}{r}=\frac{5q}{4\pi\epsilon_0 r}$.So $V$ is positive and non-zero, which rules out the option claiming $V=0$.Electric field at the centre $O$:Each charge creates at $O$ a field of the same magnitude $E_i=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$.Since all charges are positive, each field points radially away from its vertex, i.e. along the line joining that vertex to $O$.Hence the five field vectors at $O$ have equal magnitudes and their directions are equally spaced, differing by $72^\circ$ from one another (symmetry of the regular pentagon).Adding them head to tail closes a regular pentagon, so their vector sum is zero.Therefore $\vec{E}=\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4+\vec{E}_5=0$.Why the other options are wrong:$\vec{E}=\frac{5q}{4\pi\epsilon_0 r^2}\hat{r}$ wrongly treats the five fields as parallel vectors and adds their magnitudes as scalars.$\vec{E}=\frac{5\sqrt{3}q}{8\pi\epsilon_0 r^2}\hat{r}$ is not produced by this symmetric arrangement of five equal charges on a circle.The option with $V=0$ is wrong because potential is a scalar addition of five positive contributions, not a vector cancellation.Answer:$V=\frac{5q}{4\pi\epsilon_0 r}$ and $\vec{E}=0$, which is option B — the currently marked answer is correct.