First slide

Dipole

Question

Three identical dipoles are arranged as shown below. What will be the net electric field at $P (k = \frac{1}{4 π ε_{0}}) ?$

Moderate

A

$\frac{k \cdot p}{x^{3}}$
B

$\frac{2 k p}{x^{3}}$
C

$z e r o$
D

$\frac{\sqrt{2} k p}{x^{3}}$

Solution

Point P lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3.
$L e t d i p o l e m o m e n t, p_{2} = p_{3} = Q L = p a n d p_{1} = 8 Q L = 8 p$
Hence, field at P
due to dipoles 1
$E_{1} = \frac{k \cdot 8 p}{{(2 x)}^{3}} = \frac{k \cdot p}{x^{3}} (towards left)$
due to dipole 2
$E_{2} = \frac{k \cdot p}{x^{2}} (towards left)$
due to dipole 3
$E_{3} = \frac{k \cdot (2 p)}{x^{3}} (towards right)$
So net field at P will be zero.

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