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 (E) at the center O of the pentagon due to these five positive charges are: (a) V = (5q)/(4πε₀ r) and E = (5√(3)q)/(8πε₀ r²)r (b) V = (5q)/(4πε₀ r) and E = 0 (c) V = 0 and E = 0 (d) V = (5q)/(4πε₀ r) and E = (5q)/(4πε₀ r²)r

Answer: (b). Potential is a scalar: each charge is at distance r from O, so V = 5·(q)/(4πε₀ r) = (5q)/(4πε₀ r) Field is a vector: five equal fields of magnitude (q)/(4πε₀ r²) point at 72° to each other, forming a closed regular pentagon, so E = 0 So the answer is (b) V = (5q)/(4πε₀ r), E = 0.

Five positive charges each having charge q are placed at the vertices… | JEE Main 2026 (22 Jan, Shift 2)

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Electric Potential & Equipotentials

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 $(\vec{E})$ at the center $O$ of the pentagon due to these five positive charges are:

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Answer: (b)

Potential is a scalar: each charge is at distance $r$ from $O$ , so $V = 5\cdot\frac{q}{4\pi\varepsilon_0 r} = \frac{5q}{4\pi\varepsilon_0 r}$
Field is a vector: five equal fields of magnitude $\frac{q}{4\pi\varepsilon_0 r^2}$ point at $72^\circ$ to each other, forming a closed regular pentagon, so $\vec{E} = 0$
So the answer is (b) $V = \frac{5q}{4\pi\varepsilon_0 r}$ , $\vec{E} = 0$ .

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