Question

Electric charges $q,q, - 2q$ 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) $2ql$
C) $\sqrt 3 ql$
D) $4ql$

Answer 1

According to question, the three charges are placed on the three corners of the equilateral triangle ABC of side $l$ . Now, we know that dipole moment is always from negative charges to a positive charge. Now, assigning the charges to the corners of the triangle and marking the direction of dipole moment. Now, by resolving both moments, we will get the resultant dipole moment of the dipole moment.

Complete step by step solution:
Now, the diagram for the system is

Now, from the diagram, we can see that ABC is a triangle and $q,q, - 2q$ are the charges which are placed at the corners A, B and C respectively.
Now, we know that dipole moment is from negative charge to positive charge.
Now, the dipole moment is from C to A and C to B.
And we know that $- 2q$ is the charge on corner C and it is distributed as 2 charges $- q$ which forms dipole moments with A and B.
Now, $\vec P$ is the dipole moment from C to B and C to A.
Now, we know that the angle of the equilateral triangle is ${60^ \circ }$ .

Now, $\vec P$ and $\vec P$ are dipole moment at ${60^ \circ }$
So, $\vec P = ql$
Now, by taking resultant of the two moments,
${{\vec P}_{res}} = \sqrt {{P^2} + {P^2} + 2P \cdot P \cdot \cos {{60}^ \circ }} \\\ {{\vec P}_{res}} = \sqrt 3 P \\\ {{\vec P}_{res}} = \sqrt 3 ql \\\$
Hence, the correct answer is $\sqrt 3 ql$ .

Hence, the correct option is C.

Note: We know that dipole moments move from negative charge to positive charge and in the above question the corner C is having the negative charge. Now, this charge will get distributed equally to form two dipole moments with corner A and B. Now, by finding the resultant, we will get our answer.