Question

An electric dipole of moment P​ is placed normal to the lines of force of electric intensity E, then find the work done in deflecting it through an angle of \[{180^0}\] ?
A. PE
B. +2PE
C. -2PE
D. Zero

Answer 1

Hint:Before we start addressing the problem, we need to know about the electric dipole moment. It measures the separation of positive and negative electrical charges within a system, in other words, it is a measure of the system's overall polarity and its SI unit is a coulomb-meter (Cm).

Formula Used:
To find the work done of a dipole moment, we have
\[W = PE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)\]
Where, P is the electric dipole of the dipole moment, E is the electric field and \[{\theta _1}\] and \[{\theta _2}\] are the angle of rotation.

Complete step by step solution:
Consider an electric dipole of moment P​ which is placed normal to the lines of force of electric intensity E, then we need to find the work done in deflecting it through an angle of \[{180^0}\]. We know the formula to find the work done of a dipole moment,
\[W = PE\left( {\cos {\theta _1} - \cos {\theta _2}} \right)\]

When the dipole moment is perpendicular to the electric field, then \[{\theta _1} = {90^0}\] and then it is rotated through an angle of \[{180^0}\] then \[{\theta _2} = {180^0}\]. Then, the above equation will become,
\[W = PE\left( {\cos {{90}^0} - \cos {{180}^0}} \right)\]
\[ \Rightarrow W = PE\left( {0 - \left( { - 1} \right)} \right)\]
\[ \therefore W = PE\]
Therefore, the work done in deflecting it through an angle of \[{180^0}\] is PE.

Hence, option A is the correct answer.

Note:We have an alternate method to this question which is as follows. We know that the dipole moment is denoted by P. now the electric field E will be produced a torque on the dipole that is,
\[\tau = PE\sin \theta \]
Now work done from rotating it from an angle \[{\theta _1}\] to \[{\theta _2}\] is,
\[W = \int\limits_{{\theta _1}}^{{\theta _2}} {\tau d\theta } \\ \]
\[\Rightarrow W = PE\int\limits_{{\theta _1}}^{{\theta _\partial }} {\sin \theta } d\theta \\ \]
\[\Rightarrow W = - PE\left( {\cos \theta } \right)_{{\theta _1}}^{{\theta _2}} \\ \]
\[\Rightarrow W = - PE\left( {\cos {\theta _2} - \cos {\theta _1}} \right) \\ \]
\[\Rightarrow W = PE\left( {\cos {{180}^0} - \cos {{90}^0}} \right) \\ \]
\[\Rightarrow W = PE\left( {\left( { - 1} \right) - 0} \right) \\ \]
\[ \therefore W = PE\]
Therefore, the work done in deflecting it through an angle of \[{180^0}\] is PE.