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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 the line joining points $(x = 0, y = 0, z = 0)$ and $(x = a, y = a, z = 0)$
B
$qa$ along the line joining points $(x = 0, y = 0, z = 0)$ and $(x = a, y = a, z = 0)$
C
$\sqrt{2}\,qa$ along +ve x direction
D
$\sqrt{2}\,qa$ along +ve y direction
Detailed Solution
The three point charges $+q$, $-2q$ and $+q$ are placed at B $(0, a, 0)$, O $(0, 0, 0)$ and A $(a, 0, 0)$.
The charge $-2q$ at the origin can be treated as two charges $-q$ and $-q$.
So the system consists of two dipoles: one due to $-q$ at O and $+q$ at A, and the other due to $-q$ at O and $+q$ at B.
Each dipole moment has magnitude $qa$ and points from the negative to the positive charge: one along $\vec{OA}$ (+x direction) and the other along $\vec{OB}$ (+y direction).
The two moments are perpendicular, so the net dipole moment is $p_{net} = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2}\,qa$
Its direction is along OP, at an angle of $45^\circ$ with the positive x-axis, i.e. along the line joining $(0, 0, 0)$ and $(a, a, 0)$.
The charge $-2q$ at the origin can be treated as two charges $-q$ and $-q$.
So the system consists of two dipoles: one due to $-q$ at O and $+q$ at A, and the other due to $-q$ at O and $+q$ at B.
Each dipole moment has magnitude $qa$ and points from the negative to the positive charge: one along $\vec{OA}$ (+x direction) and the other along $\vec{OB}$ (+y direction).
The two moments are perpendicular, so the net dipole moment is $p_{net} = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2}\,qa$
Its direction is along OP, at an angle of $45^\circ$ with the positive x-axis, i.e. along the line joining $(0, 0, 0)$ and $(a, a, 0)$.