JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(.........\,mJ\)
- A \(17\)
- B \(18\)
- C \(16\)
- D \(13\)
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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