Ten charges are placed on the circumference of a circle of radius $R$ with constant angular separation between successive charges. Alternate charges $1, 3, 5, 7, 9$ have charge $(+q)$ each, while $2, 4, 6, 8, 10$ have charge $(-q)$ each. The potential $V$ and the electric field $E$ at the centre of the circle are respectively: (Take $V = 0$ at infinity)
Answer: (C) $V = 0;\ E = 0$
**Potential:** every charge is at distance $R$, and the total charge is $5q - 5q = 0$, so $V = 0$.
**Field:** the charges are $36^\circ$ apart. Charge $k$ and charge $k+5$ are diametrically opposite and of opposite sign, so each such pair gives a field of the same magnitude $\dfrac{2kq}{R^2}$ along that diameter. The five pair-fields point in directions that are $72^\circ$ apart (like the sides of a regular pentagon), so they add to zero.
Hence $V = 0$ and $E = 0$.
Solution by Sreeraj P, M.Sc Physics