The potential at a point P due to an electric dipole is $1.8 \times 10^{5} V$ . If P is at a distance of 50cm apart from the centre ‘O’ of the dipole and the line makes an angle $60^{0}$ with the positive side of the axial line of the dipole, the moment of the dipole is

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$10^{- 4} C - m$

$10 C - m$

$10^{- 5} C - m$

$10^{- 3} C - m$

answer is D.

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Detailed Solution

$V = \frac{P \cos θ}{4 π \in_{0} r^{2}} = \frac{9 \times 10^{9} \times P \times 1}{2 \times \frac{1}{4}} \Rightarrow 1.8 \times 10^{5} = P \times 1.8 \times 10^{10} V \Rightarrow P = 10^{- 5} C - m$

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