Determine the electric dipole moment of the system of three charges,placed on the vertices of an equilateral triangle ,as shown in the figure:

1. $\left(ql\right)\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
2. $\sqrt{3}ql\frac{\hat{j}-\hat{i}}{\sqrt{2}}$
3. $-\sqrt{3}ql\hat{j}$
4. $2ql\hat{j}$

$\text{Step 1:Breaking the Charge -2q}$

Make the dipoles out of the given system of charges by breaking up charge $-2q$ as $-q$ and $-q$ as shown in Figure

$\text{Step 2: Dipole moments of the dipole formed}$

We know that the direction of a dipole is from $-q$ to $q$.

Now we have two dipole vectors ${\overrightarrow{P}}_1$ and ${\overrightarrow{P}}_2$ as shown in Figure.

Magnitude of dipoles: $P_1=ql$, $P_2=ql$

$\therefore P_1=P_2=P=ql$

$\text{Step 3: Resultant dipole moment}$

As sine components of $P_1$ and $P_2$ would cancel each other.

Net dipole moment $=P\cos {30}^o(-\hat{j})+P\cos {30}^o(-\hat{j})$

$=-2ql\cos {30}^o\hat{j}=-\sqrt{3}ql\hat{j}$

Hence, Option (C) is correct.