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 (E⃗) at the center O of the pentagon due to these five positive charges are :

Solution

Physics sir · 4 steps

Potential is a scalar, so the five terms add up; field is a vector, so the five arrows cancel by symmetry.

1. Potential adds up. Every charge sits at the same distance r from O, so each one gives q/(4πε₀ r). Potential is a scalar with no direction, so the five equal values add straight up.

   V = 5 × q/(4πε₀ r) = 5q/(4πε₀ r)

2. Field cancels. Each charge produces a field of the same size, q/(4πε₀ r²), at O, pointing away from that charge. The five arrows are 72° apart. Placed head to tail, they form a closed regular pentagon, so their sum is zero.
3. Symmetry check. Turn the whole figure by 72° about O and it looks exactly the same. A nonzero E⃗ at O would have to turn with it, but it cannot point in a new direction while the charges have not changed. So the only possible value is E⃗ = 0.
4. So. V = 5q/(4πε₀ r) and E⃗ = 0, which is option (b).

The idea to keep. Equal charges arranged symmetrically around a point always give zero field there, because the vectors cancel. Their potentials never cancel. They just add up.

Answer from the NTA final answer key. Solution by PrepWiser.