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Potential energy of a dipole in external field - class-XII
An electric dipole of length $20cm$ having $\pm 3\times { 10 }^{ -3 }C$ charge placed at ${60}^{o}$ with respect to a uniform electric field experiences a torque of magnitude $6Nm$. The potential energy of the dipole is
Here length of dipole $2a=20cm=20\times { 10 }^{ -2 }m$, Charge $q=\pm 3\times { 10 }^{ -3 }C,\theta ={ 60 }^{ o }\quad $ and torque $\tau =6Nm$
As $\tau =pE\sin { \theta } $
or $E=\cfrac { \tau }{ p\sin { \theta } } =\cfrac { \tau }{ q(2a)\sin { \theta } } \left( \because p=q(2a) \right) $
$\therefore E=\cfrac { 6 }{ 3\times { 10 }^{ -3 }\times 20\times { 10 }^{ -2 }\times \sin { { 60 }^{ o } } } =\cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 } } N{ C }^{ -1 }$
Potential energy of dipole $U=-pE\cos{\theta}=-q(2a)E\cos{\theta}$
$=-3\times { 10 }^{ -3 }\left( 20\times { 10 }^{ -2 } \right) \cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 } } \cos { { 60 }^{ o } } =\cfrac { -3\times { 10 }^{ -5 }\times 20\times { 10 }^{ 5 } }{ 5\sqrt { 3 } \times 2 } =-2\sqrt { 3 } J\quad \quad $