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```        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```
8 years ago

```							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 60o ...
now resultant = root(p2+p2+2p2cos60o) = root3 (P)
now , P = qL so resultant = root3(qL)
```
8 years ago
```							the answer of the question :the resultant dipole moment of 3 charges placed at the vertices of the triangle is root3qL or -root3 qL
```
7 years ago
```							Explanation given by vikas is correct.Except that “Direction of dipole is from negative to positive.” then only its answer will “root(3)ql”.Otherwise if “direction of dipole moments of  both is towards C …..” it will be in negative j-direction,since it is a vector quantity.Therefore answer will become “-root3(ql)”Y/\|   A(Q)-------------------B(Q)||||             C(-2Q)||----------------------------------->X
```
3 years ago
```							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
```
2 years ago
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