JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

Six point charges are kept 60° apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is _________.

( $ϵ_{0}$ is permittivity of free space)

A
$- \frac{5 Q}{8 π ϵ _{0} R ^{2}} (\hat{i} + 3 \hat{j})$
B
$- \frac{Q}{4 π ϵ _{0} R ^{2}} (3 \hat{i} - \hat{j})$
C
$- (\frac{5 Q}{8 π ϵ _{0} R ^{2}}) (\hat{i} - 3 \hat{j})$
D
$\frac{Q}{4 π ϵ _{0} R ^{2}} (3 \hat{i} - \hat{j})$

Solution

Let $E_{0} = \frac{k Q}{R ^{2}}$ be the magnitude of the field at the centre due to any single charge of magnitude $Q$ .

Pair up diametrically opposite charges. Same-sign pairs cancel; opposite-sign pairs add to give $2 E_{0}$ pointing from the positive charge toward the centre (since the negative charge pulls toward itself, equivalently the positive pushes away, and these add constructively).

With the given arrangement, the surviving contribution comes from the two opposite-sign pairs that leave a net field along the direction making 30° below the $- \hat{i}$ axis (towards the lower-right). Resolving:

$E_{n e t} = 2 E_{0} cos 30 d e g; (- \hat{i}) + 2 E_{0} sin 30 d e g; (\hat{j}) = \frac{2 k Q}{R ^{2}} [\frac{3}{2} (- \hat{i}) + \frac{1}{2} \hat{j}]$ .

$= - \frac{Q}{4 π ϵ _{0} R ^{2}} (3 \hat{i} - \hat{j})$ .

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