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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2023
  • Medium
  • Numerical

An electric dipole of dipole moment is \(6.0 \times 10^{-6}\,Cm\) placed in a uniform electric field of \(1.5 \times 10^3\,NC ^{-1}\) in such a way that dipole moment is along electric field. The work done in rotating dipole by \(180^{\circ}\) in this field will be \(.........\,mJ\)

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Answer & Solution

Correct Answer

(B) \(18\)

Step-by-step Solution

Detailed explanation

The work done in rotating the electric dipole \(=\Delta U\)

\(= U _{ f }- U _{ i }\)

\(=\left(- pE \cos \left(180^{\circ}\right)\right)-\left(- pE \cos \left(0^{\circ}\right)\right)\)

\(= pE + pE =2 pE\)

\(=2 \times 6 \times 10^{-6} \times 1.5 \times 10^3=18\,mJ\)

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