#### Question

An electric dipole of length 1 cm, which placed with its axis making an angle of 60° with uniform electric field, experience a torque of \[6\sqrt{3} Nm\] . Calculate the potential energy of the dipole if it has charge ±2 nC.

#### Solution

Torque,

\[\tau = \text { PEsin }\theta = \left( Ql \right)\text { Esin }\theta\]

Here,

Potential energy,

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

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

Dividing (2) by (1):

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

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

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

\[ \Rightarrow U = - \frac{6\sqrt{3}}{\sqrt{3}}\]

\[ \Rightarrow U = - 6 J\]

Is there an error in this question or solution?

Solution An electric dipole of length 1 cm, which placed with its axis making an angle of 60° with uniform electric field, experience a torque of 6 √ 3 N m . Calculate the potential energy Concept: Electric Dipole.