Question

Three charged particle A, B and C with charges \(-4q\), \(2q\) and \(-2q\) are present on the circumference of a circle of radius \(d\). The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is :

• A. \(\frac{3\sqrt{3}\,q}{4\pi\varepsilon_0 d^2}\)
• B. \(\frac{\sqrt{3}\,q}{4\pi\varepsilon_0 d^2}\)
• C. \(\frac{\sqrt{3}\,q}{\pi\varepsilon_0 d^2}\)
• D. \(\frac{2\sqrt{3}\,q}{\pi\varepsilon_0 d^2}\)

Hint:

Add the x-components of the three fields at O (each of magnitude \(kQ/d^2\)); the negative charges A and C produce fields directed towards them, the positive charge B produces a field directed away from it.

Answer 1

The Correct Option is C

Let \(k=\frac{1}{4\pi\varepsilon_0}\). Each charge is at distance \(d\) from O.
A (\(-4q\)) lies at \(30^\circ\) above the +x axis; its field at O points towards A: magnitude \(\frac{4kq}{d^2}\), x-component \(\frac{4kq}{d^2}\cos30^\circ=\frac{2\sqrt{3}kq}{d^2}\).
C (\(-2q\)) lies at \(30^\circ\) below the +x axis; its field points towards C: x-component \(\frac{2kq}{d^2}\cos30^\circ=\frac{\sqrt{3}kq}{d^2}\).
B (\(+2q\)) is the mirror image of A about the y-axis (line OB makes \(60^\circ\) with the y-axis, i.e. \(30^\circ\) with the -x axis); its field points away from B, i.e. towards the +x side: x-component \(\frac{2kq}{d^2}\cos30^\circ=\frac{\sqrt{3}kq}{d^2}\).
Total: \(E_x=\frac{kq}{d^2}\left(2\sqrt{3}+\sqrt{3}+\sqrt{3}\right)=\frac{4\sqrt{3}kq}{d^2}=\frac{4\sqrt{3}\,q}{4\pi\varepsilon_0 d^2}=\frac{\sqrt{3}\,q}{\pi\varepsilon_0 d^2}\).
Answer: (3) \(\frac{\sqrt{3}\,q}{\pi\varepsilon_0 d^2}\)