Question

Two point-charges of \[+1.0\times {{10}^{-6}}\text{ C}\]and \[-1.0\times {{10}^{-6}}\text{ C}\] are at a distance \[\text{2}\text{.0 cm}\] apart from each other. This electric dipole is situated in a uniform electric field of \[1.0\times {{10}^{5}}\text{ V-}{{\text{m}}^{-1}}\]. What will be the maximum torque acting on it due to the field? What will be the work done for rotating it through 180 from the equilibrium position?

Answer 1

Hint: When the electric dipole is placed perpendicular to the electric field, then the torque acting on it will be maximum. On integrating the formula of the torque acting on the dipole, the work done in rotating the dipole through angle \[\theta \] from the direction of the electric field E is obtained.

Formula used: The magnitude of dipole moment, p, of an electric dipole is given by
\[p=2ql\]
Where q is the charge and 2l is the distance between the charges.
The magnitude of torque,\[\tau \], acting upon a dipole of moment \[\overrightarrow{p}\] placed in a uniform electric field \[\overrightarrow{E}\] making an angle \[\theta \] is given by:
\[\tau =pE\sin \theta \]
Work done in rotating the dipole through angle \[\theta \] from the direction of the electric field E is given by:
\[W=pE(1-\cos \theta )\]

Complete step by step answer:
The charge at one end of the dipole , \[q=1.0\times {{10}^{-6}}\text{ C}\]
The distance between the two charges, \[2l=2.0\text{ cm = 0}\text{.02 m}\]
The magnitude of electric field, \[E=1.0\times {{10}^{5}}\text{ V-}{{\text{m}}^{-1}}\]
The electric dipole moment p is,
\[\begin{align}
  & p=2ql=(1.0\times {{10}^{-6}}\text{ C})(\text{0}\text{.02 m)} \\
 &\Rightarrow p=2\times {{10}^{-8}}\text{ Cm} \\
\end{align}\]
The torque acting on the dipole due to an electric field \[E=1.0\times {{10}^{5}}\text{ V-}{{\text{m}}^{-1}}\] is
\[\tau =pE\sin \theta \]
The torque acting on the dipole is maximum when \[\theta ={{90}^{\text{o}}}\], so
\[\begin{align}
  & {{\tau }_{\max }}=(2\times {{10}^{-8}}\text{ Cm)(1}\text{.0}\times \text{1}{{\text{0}}^{5}}\text{ V-}{{\text{m}}^{-1}})(\sin {{90}^{\text{o}}})\text{ } \\
 &\Rightarrow {{\tau }_{\max }}=2\times {{10}^{-3}}\text{ Nm }\!\![\!\!\text{ sin9}{{\text{0}}^{\text{o}}}=1] \\
\end{align}\]
Now, in equilibrium position,\[\theta =0\]. So work done in rotating the dipole through \[{{180}^{\text{o}}}\] from the equilibrium position is:
\[\begin{align}
  & W=pE(1-\cos \theta ) \\
 &\Rightarrow W=(2\times {{10}^{-8}}\text{ Cm})(1.0\times {{10}^{5}}\text{ V-}{{\text{m}}^{-1}})(1-\cos {{180}^{\text{o}}})\text{ } \\
 &\Rightarrow W=(2\times {{10}^{-8}}\text{ Cm})(1.0\times {{10}^{5}}\text{ V-}{{\text{m}}^{-1}})(2)\text{ }\!\![\!\!\text{ }\cos {{180}^{\text{o}}}=-1] \\
 &\Rightarrow W=4\times {{10}^{-3}}\text{ J} \\
\end{align}\]

Note: The torque acting on an electric dipole always tends to align the dipole in the direction of the electric field.
In a uniform electric field, a dipole feels a torque but no net force. However, if a dipole is placed in a non-uniform electric field, it experiences a net force in addition to a torque.