Question

Three point charges $+q$, $-2q$ and $+q$ are placed at $(x=0,y=a,z=0)$, $(x=0,y=0,z=0)$ and $(x=a,y=0,z=0)$ respectively. The magnitude and direction of the electric dipole moment vector for the given system of charges are

• A. $\sqrt{2}qa$ along $+x$ direction
• B. $\sqrt{2}qa$ along $+y$ direction
• C. $\sqrt{2}qa$ along the line joining points $(x=0,y=0,z=0)$ and $(x=a,y=a,z=0)$
• D. $qa$ along the line joining points $(x=a,y=0,z=0)$ and $(x=a,y=a,z=0)$

Answer 1

The correct option is C $\sqrt{2}qa$ along the line joining points $(x=0,y=0,z=0)$ and $(x=a,y=a,z=0)$

The position of three point charges is shown in figure given below,


The $-2q$ charge placed at origin can be visualised as two $-q$ charges placed at the same point.

Thus we have two dipole moments i.e $\overrightarrow{p_1}$ and $\overrightarrow{p_2}$ respectively along $+ve-y$ and $+ve-x$ axis respectively.


The magnitude of each dipole moment is equal.
$\Rightarrow |{\overrightarrow{p}}_1|=|{\overrightarrow{p}}_2|=q\times a=qa$

Using superposition principle, the net dipole moment of system is given by:

${\overrightarrow{p}}_{\text{net}}={\overrightarrow{p}}_1+{\overrightarrow{p}}_2$

${\overrightarrow{p}}_{\text{net}}=qa\hat{i}+qa\hat{j}....\left(1\right)$

$\Rightarrow |{\overrightarrow{p}}_{\text{net}}|=\sqrt{(qa{)}^2+(qa{)}^2}=\sqrt{2q^2{a}^2}$

$\therefore |{\overrightarrow{p}}_{\text{net}}|=\sqrt{2}qa$

From Eq.$\left(1\right)$ the angle made by resultant vector is,

$\tan \theta =\frac{qa}{qa}=1$

$\theta ={45}^{\circ }$

Therefore the ${\overrightarrow{p}}_{\text{net}}$ is directed along the straight line $y=x$ which is joining points $(0,0,0)$ and $(a,a,0)$

$\therefore$ Option$\left(3\right)$ is the correct answer.

Why this question? Tip: The net dipole moment vector is along the straight line $y=x$, because the $x\&y$ components of resultant dipole moment vector are equal.