# 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 - Physics

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, 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):

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

$\Rightarrow U = - \frac{\tau}{\tan\theta}$

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

$\Rightarrow U = - \frac{6\sqrt{3}}{\sqrt{3}}$

$\Rightarrow U = - 6 J$

Concept: Electric Dipole
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