Electrostatics

JEE Main level, single correct, medium

262 of 311

Question 262 of 311

An electric field of 1000 V/m1000 \mathrm{~V} / \mathrm{m} is applied to an electric dipole at angle of 45∘45^{\circ}. The value of electric dipole moment is 10−29C⋅m.10^{-29} \mathrm{C} \cdot \mathrm{m} . What is the potential energy of the electric dipole?

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Answer: (B) −7×10−27 J-7 \times 10^{-27} \mathrm{~J}

Potential energy of a dipole is given by U=−P→E→\mathrm{U}=-\overrightarrow{\mathrm{P}} \overrightarrow{\mathrm{E}}

=−PEcos⁡θ=-P E \cos \theta

[Where θ=\theta= angle between dipole and perpendicular to the field =−(10−29)(103)cos⁡45∘=-\left(10^{-29}\right)\left(10^{3}\right) \cos 45^{\circ}

=−0.707×10−26 J=−7×10−27 J=-0.707 \times 10^{-26} \mathrm{~J}=-7 \times 10^{-27} \mathrm{~J}

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