Question #70bb8

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Question #70bb8

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Answer 1 · 2016-12-25T08:56:51

So option (1) is correct

Explanation:

It is obvious from the given figure that three charge points are situated at thrre vertices of an isosceles triangle in which two equal sides of length $l$ $l$ is inclined at an angle $120^{\circ}$ $120^@$ and $- 2 q$ $-2q$ charge resides at the point of intersection of these two sides. So each of the other two angles of the isosceles triangle will be $30^{\circ}$ $30^@$ .The c.g of two $+ q$ $+q$ charge will be at the mid point of the line segment joining other two vertices of the triangle and the distance of resultatant $+ 2 q$ $+2q$ charge at this middle point from $- 2 q$ $-2q$ charge will be $d = l \times {sin 30}^{\circ} = \frac{l}{2}$ $d=lxxsin30^@=l/2$

Hence the dipolemoment of the system is given by

$μ = 2 q \times d = 2 q \times \frac{l}{2} = q l$ $mu=2qxxd=2qxxl/2=ql$

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Question #70bb8

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