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CBSE Class 12 Physics 2014 Delhi Set 3 Paper

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Question : 4 of 8
Marks: +1, -0
An electric dipole of length 2 cm2\text{ cm} , when placed with its axis making an angle of 6060^{\circ} with a uniform electric field, experiences torque of 63 Nm6\sqrt{3}\text{ Nm} . Calculate the potential energy of the dipole, if it has a charge of ±2n C\pm 2n\text{ C} .
Solution:  
τ=pEsinθ\tau = pE \sin\theta
63=pEsinθ6\sqrt{3}=pE \sin\theta
pE32=63pE\frac{\sqrt{3}}{2}=6\sqrt{3}
So, pE=12pE=12
Potential energy of the dipole =pEcosθ=-pE\cos\theta
U=pEcos60U=-pE\cos 60^{\circ}
U=12×0.5U=-12\times 0.5
U=6 JU=-6\text{ J}
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