Question

A dipole is placed parallel to the electric field. If \[W\] is the work done in rotating the dipole by 60, then the work done in rotating it by 180 is:
A. \[2W\]
B. \[3W\]
C. \[4W\]
D. \[W/2\]

Answer 1

Hint:Use the formula for work done in rotating a dipole placed in an electric field. This formula gives the relation between the dipole moment, electric field, initial angle between the dipole and electric field and final angle between the dipole and electric field. First determine the value of work done in rotating the dipole from parallel position to \[60^\circ \]. Then determine the work done in rotating the dipole from \[60^\circ \] to \[180^\circ \] in terms of initial work done.

Formula used:
The work done in rotating the dipole placed in the electric field is
\[W = PE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)\] …… (1)
Here, \[P\] is the dipole moment, \[E\] is the electric field, \[{\theta _1}\] is the initial angle of dipole with electric field and \[{\theta _2}\] is the final angle of dipole moment with electric field.

Complete step by step answer:
We have given that initially the dipole is placed parallel to the electric field. This means that the initial angle between the dipole and electric field is \[0^\circ \].
\[{\theta _1} = 0^\circ \]
The work done in rotating the dipole from the initial parallel position to \[60^\circ \] is \[W\].
\[{\theta _2} = 60^\circ \]
The work done in rotating the dipole from the initial parallel position to \[60^\circ \] is given by equation (1). Substitute \[0^\circ \] for \[{\theta _1}\] and \[60^\circ \] for \[{\theta _2}\] in equation (1).
\[W = PE\left( {\cos 0^\circ - \cos 60^\circ } \right)\]
\[ \Rightarrow W = PE\left( {1 - \dfrac{1}{2}} \right)\]
\[ \Rightarrow W = \dfrac{{PE}}{2}\]

Let \[W'\] be the work done in rotating the dipole from \[60^\circ \] to \[180^\circ \].Here, the initial angle between the dipole and electric field is \[60^\circ \] and the final angle between the dipole and electric field is \[180^\circ \].
\[{\theta _1} = 60^\circ \]
\[{\theta _2} = 180^\circ \]
Substitute \[W'\] for \[W\], \[60^\circ \] for \[{\theta _1}\] and \[180^\circ \] for \[{\theta _2}\] in equation (1).
\[W' = PE\left( {\cos 60^\circ - \cos 180^\circ } \right)\]
\[ \Rightarrow W' = PE\left( {\dfrac{1}{2} - \left( { - 1} \right)} \right)\]
\[ \Rightarrow W' = \dfrac{{3PE}}{2}\]
Substitute \[W\] for \[\dfrac{{PE}}{2}\] in the above equation.
\[ \therefore W' = 3W\]
Therefore, the work done in rotating the dipole to \[180^\circ \] is \[3W\].

Hence, the correct option is B.

Note:One can also solve the same question by another method. One can determine the work done in rotating the dipole from parallel position to \[60^\circ \] and then work done in rotating dipole from parallel position to \[180^\circ \]. Then subtract the work done in rotating dipole to \[180^\circ \] from work done in rotating dipole to \[60^\circ \].