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An Electric Dipole of Length 2 Cm, When Placed with Its Axis Making an Angle of 60° with a Uniform Electric Field, Experiences a Torque of 8 √ 3 Nm. Calculate the Potential Energy of the Dipole

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Question

An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of \[8\sqrt{3}\] Nm. Calculate the potential energy of the dipole, if it has a charge \[\pm\] 4 nC.

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Solution

Torque, \[\tau = \text { PE sin }\theta = \left( Ql \right)\text { E sin }\theta\]  ........ (1)

Here, l is the length of the dipole, Q is the charge and E is the electric field.
Potential energy,

\[\text { U = - P Ecos }\theta = - \left( Ql \right)\text {Ecos}\theta\]

Dividing (2) by (1):

\[\frac{\tau}{U} = \frac{\left( Ql \right)E\sin\theta}{- \left( Ql \right)E\cos\theta} = - \tan\theta\]

\[ \Rightarrow U = - \frac{\tau}{\tan\theta}\]

\[ \Rightarrow U = - \frac{\tau}{\tan {60}^o}\]

\[ \Rightarrow U = - \frac{8\sqrt{3}}{\sqrt{3}}\]

\[ \Rightarrow U = - 8 J\]

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2013-2014 (March) Delhi Set 2

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