Question

An electric dipole of moment p is kept along an electric field E. The work done in rotating it from an equilibrium position by an angle theta is

Answer 1

Answer:

We know that,

$\boxed{\rm \tau=pEsin \theta }$

Electric field will be produced a torque on the dipole

According to the question,

We are asked to find the work done in rotating it from an equilibrium position by angle $\rm \theta$ is given as

Work done (W) =

Using integration,

We get,

$\implies \rm W=\int\limits^\theta_0 {\tau\; d \theta}$

$\implies \rm W=\int\limits^\theta_0 {pEsin \theta\; d \theta}$

$\implies \rm W=pE\int\limits^\theta_0 {sin \theta\;d \theta}$

$\implies \rm W=-pE(cos \theta-cos0)$

$\implies \rm W=-pE(cos \theta-1)$

$\implies \rm W=pE(1-cos \theta)$

$\boxed{\boxed{\rm W=pE(1-cos \theta)}}$

Hence,

The work done in rotating it from an equilibrium position by an angle theta is given as

W = pE(1 - cosθ)

Answer 2

$\huge\mathtt{{\colorbox{silver} {ANSWER~}}}$

We know that,

$\boxed{\rm \tau=pEsin \theta }$

Electric field will be produced a torque on the dipole

According to the question,

We are asked to find the work done in rotating it from an equilibrium position by angle $\rm \theta$ is given as

Work done (W) =

Using integration,

We get,

$\implies \rm W=\int\limits^\theta_0 {\tau\; d \theta}$

$\implies \rm W=\int\limits^\theta_0 {pEsin \theta\; d \theta}$

$\implies \rm W=pE\int\limits^\theta_0 {sin \theta\;d \theta}$

$\implies \rm W=-pE(cos \theta-cos0)$

$\implies \rm W=-pE(cos \theta-1)$

$\implies \rm W=pE(1-cos \theta)$

$\boxed{\boxed{\rm W=pE(1-cos \theta)}}$

Hence,

The work done in rotating it from an equilibrium position by an angle theta is given as

W = pE(1 - cosθ)