Three point charges +q,-2q and +q are placed | vedclass

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Question

(B) The given charge assembly can be represented using the coordinate axes $x$ and $y$ as shown in the figure.The charge $-2q$ is placed at the origin

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

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(B) The given charge assembly can be represented using the coordinate axes $x$ and $y$ as shown in the figure.The charge $-2q$ is placed at the origin $O(0, 0, 0)$. One $+q$ charge is placed at $(a, 0, 0)$ and the other $+q$ charge is placed at $(0, a, 0)$.This system can be viewed as two electric dipoles: one along the $x$-axis with dipole moment $\vec{p}_1 = q a \hat{i}$ and another along the $y$-axis with dipole moment $\vec{p}_2 = q a \hat{j}$.The resultant dipole moment $\vec{P}_R$ is the vector sum of these two dipoles:$\vec{P}_R = \vec{p}_1 + \vec{p}_2 = qa \hat{i} + qa \hat{j}$.The magnitude of the resultant dipole moment is:$P_R = \sqrt{(qa)^2 + (qa)^2} = \sqrt{2} qa$.The direction of the resultant dipole moment is along the vector $\hat{i} + \hat{j}$,which is the line joining the origin $(0, 0, 0)$ and the point $(a, a, 0)$.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

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

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