What work must be done to rotate an electric dipole through an angle $\theta $ with the electric field, if an electric dipole of moment $p$ is placed in a uniform electric field $E$ with $p$ parallel to $E$?
Answer
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Hint: A dipole is a pair of two electric charges of equal magnitude and opposite signs. To solve the problem, apply the formula for external work required to rotate the dipole in the given external electric field $E$ i.e., $dW = \tau d\theta $.
$\tau $ is the external torque acting on the dipole and it is given by, $\vec \tau = \vec p \times \vec E$
$p$ is the dipole moment.
Complete step by step answer:
The given dipole is placed in a uniform electric field $\vec E$.
The torque on a dipole is given by $\overrightarrow \tau = \overrightarrow p \times \overrightarrow E $.
$\tau = pE\sin \theta $
Where, $\theta $ is the angle between the dipole moment $p$ and the electric field $E$.
The work to be done to rotate the dipole by an angle $d\theta $ is $dW = \tau d\theta $
The total work done to rotate the dipole from ${\theta _1}$ to ${\theta _2}$is $W = \int_{{\theta _1}}^{{\theta _2}} {dW} $
$W = \int_{{\theta _1}}^{{\theta _2}} {\tau d\theta } $
Substitute the magnitude of torque $\tau $
$ \Rightarrow W = \int_{{\theta _1}}^{{\theta _2}} {pE\sin \theta d\theta } $
Integrate the above equation
$ \Rightarrow W = pE\left[ { - \cos \theta } \right]_{_{{\theta _1}}}^{{\theta _2}}$
$ \Rightarrow W = pE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
It is given that initially the dipole is placed parallel to the external electric field $E$.
${\theta _1} = {0^0}$ and later rotated with ${\theta _2} = \theta $
Therefore, Total work done $W = pE\left( {\cos {0^0} - \cos \theta } \right)$
Or $W = pE\left( {1 - \cos \theta } \right)$
Hence, the correct option is (A) $pE\left( {1 - \cos \theta } \right)$.
Note: The total external force acting on a dipole in an external electric field is always zero. The external torque acting on a dipole is zero when the dipole moment is either parallel or antiparallel with the external electric field.
The total work done by a system is equal to the potential energy stored in the system.
$\tau $ is the external torque acting on the dipole and it is given by, $\vec \tau = \vec p \times \vec E$
$p$ is the dipole moment.
Complete step by step answer:
The given dipole is placed in a uniform electric field $\vec E$.
The torque on a dipole is given by $\overrightarrow \tau = \overrightarrow p \times \overrightarrow E $.
$\tau = pE\sin \theta $
Where, $\theta $ is the angle between the dipole moment $p$ and the electric field $E$.
The work to be done to rotate the dipole by an angle $d\theta $ is $dW = \tau d\theta $
The total work done to rotate the dipole from ${\theta _1}$ to ${\theta _2}$is $W = \int_{{\theta _1}}^{{\theta _2}} {dW} $
$W = \int_{{\theta _1}}^{{\theta _2}} {\tau d\theta } $
Substitute the magnitude of torque $\tau $
$ \Rightarrow W = \int_{{\theta _1}}^{{\theta _2}} {pE\sin \theta d\theta } $
Integrate the above equation
$ \Rightarrow W = pE\left[ { - \cos \theta } \right]_{_{{\theta _1}}}^{{\theta _2}}$
$ \Rightarrow W = pE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
It is given that initially the dipole is placed parallel to the external electric field $E$.
${\theta _1} = {0^0}$ and later rotated with ${\theta _2} = \theta $
Therefore, Total work done $W = pE\left( {\cos {0^0} - \cos \theta } \right)$
Or $W = pE\left( {1 - \cos \theta } \right)$
Hence, the correct option is (A) $pE\left( {1 - \cos \theta } \right)$.
Note: The total external force acting on a dipole in an external electric field is always zero. The external torque acting on a dipole is zero when the dipole moment is either parallel or antiparallel with the external electric field.
The total work done by a system is equal to the potential energy stored in the system.
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