Question

Three point charges $+q,-2q$ and $+q$ are placed at points $(x=0,y=a,z=0)$, $(x=0,y=0,z=0)$ and $(x=a,y=0,$ $z=0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly 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,$$y=a,z=0)$
• D. $qa$ along the line joining points
  $(x=0,y=0,z=0)$ and $(x=a,$$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 diagram for the given problem:


Now, the charge $-2q$ is divided to $-q$ and $-q$ to form a dipole as shown in the figure,


Dipole moment for both the dipoles is $P=qa$.


Now. net dipole moment is

$P_{net}=\sqrt{P^2+P^2}$

$\Rightarrow P_{net}=\sqrt{{\left(qa\right)}^2+{\left(qa\right)}^2}$

$\therefore P_{net}=\sqrt{2}qa$

From the figure, $P_{net}$ is along $y=x$ that means, along the line joining $(0,0,0),$ $(a,a,0)$.

Hence, option (c) is the correct answer.