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CBSE Class 12 Physics 2014 Delhi Set 2 Paper

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Question : 3 of 8
Marks: +1, -0
An electric dipole of length 2 cm2\ \mathrm{cm}, when placed with its axis making an angle of 60∘60^{\circ} with a uniform electric field, experiences a torque of 83 Nm8\sqrt{3}\ \mathrm{Nm}. Calculate the potential energy of the dipole, if it has a charge of ±4 nC\pm 4\ \mathrm{nC}.
Solution:  
τ  =pE  sin⁡θ\tau\;=pE\;\sin\theta
83  =pE  sin⁡θ8\sqrt{3}\;=pE\;\sin\theta
pE  32  =83pE\;\frac{\sqrt{3}}{2}\;=8\sqrt{3}
So, pE=16pE=16
Potential energy of the dipole
  =−pEcos⁡θ\;=-pE\cos\theta
U  =−pEcos⁡60U\;=-pE\cos 60
U  =−16×0.5U\;=-16 \times 0.5
U  =−8 JU\;=-8\ \mathrm{J}
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