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Q.

The potential at a point P due to an electric dipole is 1.8×105V . If P is at a distance of 50cm apart from the centre ‘O’ of the dipole and the line makes an angle 600 with the positive side of the axial line of the dipole, the moment of the dipole is

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a

10 C–m

b

10−3C−m

c

10−4C−m

d

10−5C−m

answer is D.

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Detailed Solution

V=Pcosθ4π∈0r2=9×109×P×12×14 ⇒1.8×105=P×1.8×1010V ⇒P=10-5C-m
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