Question

Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:
(A) √3 ql. (jˆ - î)/√2
(B) ql. (jˆ - î)/√2
(C) 2ql jˆ (D) -√3 ql jˆ

Answer 1

Given :

Equilateral triangle having three charges +q, +q and -2q.

Length of side = l

To find :

Electric dipole moment.

Solution :

Firstly consider -2q charge as two bodies of charge -q each.

This division of charges can be considered as two dipoles

One dipole is from upper point to origin.

Second dipole is from upper point to x.

As the dipole moment P :

$P = q \cdot(length)$

So,

First dipole moment

$P_1 = q \cdot(l)$

in the direction of origin.

Second dipole moment

$P_2 = q \cdot(l)$

in the direction of x.

magnitude of both dipole moment is same.

So, when we make components of both the dipoles. The angle between vertical line and direction of dipole is 30°

Horizontal components of both dipoles :

$P_{1 h } = q l \sin30 \degree ( - \hat i) \\ P_{2 h } = q l \sin30 \degree ( \hat i)$

So,

Total horizontal component :

$P_{ h } = P_{ 1h } + P_{2 h } \\ = q l \sin30 \degree ( - \hat i) + q l \sin30 \degree ( \hat i) \\ = 0$

Vertical components of both dipoles :

$P_{1 v } = q l \cos30 \degree ( - \hat j) \\ P_{2 v} = q l \cos30 \degree ( - \hat j)$

So,

Total vertical component :

$P_{ v } = P_{ 1v } + P_{2 v} \\ = q l \cos30 \degree ( - \hat j) + q l \cos30 \degree ( - \hat j) \\ = - 2ql \cos30 \degree \hat j \\ = - \sqrt{3} \: \: ql \hat j$

So,

Total dipole moment of the system :

$P = P_{ h } + P_{v } \\ P = 0 \: - \sqrt{3} \: \: ql \hat j \\ P = - \sqrt{3} \: \: ql \hat j \:$

Option (D)