charges q,q,-2q are placed at the corners of equilateral triangle of side l. what is the net dipole moment?

Ans root3 ql.

pl explain

let A,B,C are three vertices ...

charge at A,B is q but at C charge is -2q ...

A(q) ...................B(q)

           C(-2q)

charge of C can be written as -q + -q

q charge of A & -q charge of C becomes 1 dipole ...

q charge of B & another -q charge of C makes another dipole ..

direction of dipole moments of  both is rowards C .....

angle bw them will be 60^o ...

now resultant = root(p²+p²+2p²cos60^o) = root3 (P)

now , P = qL so resultant = root3(qL)

the answer of the question :the resultant dipole moment of 3 charges placed at the vertices of the triangle is root3qL or -root3 qL

Dipole is only an applicable condition when and only when the charges are also equal in magnitude rather than only being opposite.So, 1st we break -2q charge at C as -q +(-q).Now there are 2 -q charges at C i.e. 1 to complete dipole with A and another with B.Now,Dipole due AC=dipole due BC = q.aSo net dipole = vector sum dipole due AC and due BC where angle b\w the 2 vectors =60°So net dipole=√(qa)²+(qa)²+2qacos60° =√3qa

Dear Student

Consider an equilateral triangle of side l. Point A and B have charges +q where A has -2q charge. The system as two dipoles with P₁ and P₂ along BA and CA respectively. Hence, the magnitude of P₁ and P₂ are equal to (ql) and the angle between them is 60°. Therefore, the resultant of two is the net dipole moment.
Now resultant = √(p²+p²+2p²cos60^o) = (P)√3

now , P = qL
so resultant = qL√3

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya