JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

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 $E = 0$
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 $E = 0$

Solution

Each vertex is at the same distance $r$ from the centre O. Potential is a scalar that adds directly:

$V = 5 \times \frac{k q}{r} = \frac{5 q}{4 π ϵ _{0} r}$ .

For the field, by the symmetry of the regular pentagon, the five equal field vectors point radially outward at angles separated by 72°. Their vector sum is zero. So $E = 0$ .

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