Q.
Three identical dipoles are arranged as shown below. What will be the net electric field at $P\left( k = \frac{1}{4\pi\varepsilon_{0}} \right)$
A) $\frac{k.p}{x^{3}}$
B) $\frac{2kp}{x^{3}}$
C) Zero
D) $\frac{\sqrt{2}kp}{x^{3}}$
Explanation
(c) Point $P$ lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3
[image:1_image247.emf]
Hence field at $P$
due to dipole 1 $E_{1} = \frac{k.p}{x^{3}}$ [towards left]
due to dipole 2
$E_{2} = \frac{k.p}{x^{2}}$ [towards left]
Due to dipole 3 $E_{3} = \frac{k.(2p)}{x^{3}}$ [towards right]
So net field at $P$ will be zero
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.