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