Electrostatics

JEE Main level, single correct, medium

59 of 311

Question 59 of 311

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)

Show answer and solution

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

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.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electric field lines simulationDrag the sliders and watch the numbers in this question change.Revise Electrostatics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Electrostatics (Electric Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.