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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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detailed solution

Correct option is D

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

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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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Ready to Test Your Skills?

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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