Question

Electric charge $q$ , $q$ and $-2q$ are placed at the corners of an equilateral triangle $ABC$ of side $L$. The magnitude of electric dipole moment of the system is

• A. $qL$
  
• B. $\sqrt{2}qL$
  
• C. $\sqrt{3}qL$
  
• D. $4qL$
  

Answer 1

The correct option is C $\sqrt{3}qL$

$\text{Step 1: Breaking the charges}$

In the given situation, we do not se any dipole directly. But we can make dipoles by breaking the charges.

So, We Break the $-2q$ charge in two $-q$ charges, and we get two dipoles as shown in the figure.

Dipole moment of both dipoles $P=qL$

$\text{Step 2: Calculation of net dipole moment}$

Since dipole is vector quantity, so we add the two dipoles vectorially to get the resultant dipole moment.

Both dipoles are at angle $\theta ={60}^{\circ }$

So, Resultant $P_{\text{net}}=\sqrt{P^2+P^2+2\left(P\right)\left(P\right)\cos 60}$

$=\sqrt{3P^2}$ $=\sqrt{3}P$

$P_{\text{net}}=\sqrt{3}(q\times L)$

Hence, Option $\left(C\right)$ is correct.