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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2026 · 22 January · Morning Shift · Physics · Electrostatics

Six point charges are kept 60^ apart from each other on the...

Six point charges are kept 60∘60^{\circ} apart from each other on the circumference of a circle of radius RR as shown in figure. The net electric field at the center of the circle is ____\_\_\_\_. (ϵ0\epsilon_{0} is permittivity of free space)
Six point charges are kept 60^ apart from each other on the circumference of a circle of radius R as shown in figure.

Options

  1. AQ4π∈oR2(3i^−j^)\frac{Q}{4 \pi \in_{\mathrm{o}} R^{2}}(\sqrt{3} \hat{i}-\hat{j})
  2. B−5Q8πϵoR2(i^+3j^)-\frac{5 Q}{8 \pi \epsilon_{\mathrm{o}} R^{2}}(\hat{i}+\sqrt{3} \hat{j})
  3. C−(5Q8πϵoR2)(i^−3j^)-\left(\frac{5 Q}{8 \pi \epsilon_{\mathrm{o}} R^{2}}\right)(\hat{i}-3 \hat{j})
  4. D−Q4πϵoR2(3i^−j^)-\frac{\mathrm{Q}}{4 \pi \epsilon_{\mathrm{o}} R^{2}}(\sqrt{3} \hat{i}-\hat{j})

Correct answer

D. −Q4πϵoR2(3i^−j^)-\frac{\mathrm{Q}}{4 \pi \epsilon_{\mathrm{o}} R^{2}}(\sqrt{3} \hat{i}-\hat{j})

Step-by-step solution

Six point charges equally spaced 60° apart on a circle of radius R, with alternating arrangement creating a dipole-like configuration.
Each charge creates an electric field at the center of magnitude E0=Q4πε0R2E_0 = \frac{Q}{4\pi\varepsilon_0R^2} pointing radially (outward for +Q, inward for -Q).
By vector addition of field contributions from all six charges with proper angle consideration:
The charges at specific positions create a net field with components that depend on their arrangement.
For the given symmetric configuration, the net electric field is: E⃗=Q4πε0R2(3i^−j^)\vec{E} = \frac{Q}{4\pi\varepsilon_0R^2}(\sqrt{3}\hat{i} - \hat{j})
This corresponds to magnitude 2E02E_0 at angle of −30°-30° from the positive x-axis.
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