Physics · Electrostatics

JEE Main 2026 — 22 January, Morning Shift — Question 27

Six point charges are kept $60^{\circ}$ 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 $\_\_\_\_$ . ( $\epsilon_{\mathrm{o}}$ is permittivity of free space)

Option A:
$-\frac{5 Q}{8 \pi \epsilon_{o} R^{2}}(\hat{i}+\sqrt{3} \hat{j})$
Option B:
$-\frac{\mathrm{Q}}{4 \pi \epsilon_{\mathrm{o}} \mathrm{R}^{2}}(\sqrt{3} \hat{\mathrm{i}}-\hat{\mathrm{j}})$
Correct
Option C:
$-\left(\frac{5 Q}{8 \pi \epsilon_{o} R^{2}}\right)(\hat{i}-3 \hat{j})$
Option D:
$\frac{\mathrm{Q}}{4 \pi \in_{\mathrm{o}} \mathrm{R}^{2}}(\sqrt{3} \hat{\mathrm{i}}-\hat{\mathrm{j}})$

Answer: B

Step-by-step solution

Let $\frac{\mathrm{KQ}}{\mathrm{r}^{2}}=\mathrm{E}_{0}$ $\overrightarrow{\mathrm{E}}_{\text {net }}=2 \mathrm{E}_{0} \cos 30^{\circ}(-\hat{\mathrm{i}})+2 \mathrm{E}_{0} \sin 30^{\circ}(\hat{\mathrm{j}})$ $=\frac{2 \mathrm{kQ}}{\mathrm{r}^{2}}\left[\frac{\sqrt{3}}{2}(-\hat{\mathrm{i}})+\frac{1}{2} \hat{\mathrm{j}}\right]$ $=\frac{-1 Q}{4 \pi \varepsilon_{0} r^{2}}(\sqrt{3} \hat{i}-\hat{j})$

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Properties of Electric Charge and Coulomb's Law

Electric field in a region is given by vecE=A x hati+B y hatj , where A=10 V / m^2 and B=5 V / m^2 . If the electric potential at a point…

Answer: B

Step-by-step solution

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