JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Electrostatics

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) $V = \frac{5 q}{4 π ε _{0} r}$ and $E = 0$ ✓ Correct
(B) $V = \frac{5 q}{4 π ε _{0} r}$ and $E = \frac{5 3 q}{8 π ε _{0} r ^{2}} \overset{r}{^}$
(C) $V = \frac{5 q}{4 π ε _{0} r}$ and $E = \frac{5 q}{4 π ε _{0} r ^{2}} \overset{r}{^}$
(D) $V = 0$ and $E = 0$

Answer: (A)

Solution

Potential is a scalar, so the five charges add directly:

V = 5 \cdot \frac{1}{4 π ε _{0}} \frac{q}{r} = \frac{5 q}{4 π ε _{0} r}

. The field is a vector; the five equal charges occupy the vertices of a regular pentagon, so the five field contributions at the centre are equal in magnitude and separated by

7 2^{\circ}

, and their vector sum is zero. Thus

V = \frac{5 q}{4 π ε _{0} r}

and

E = 0

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Electrostatics

Solution

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