A molecule with a dipole moment p is placed in the electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field, the work required to be done by an external agency is
A. \[ - 2pE\]
B. \[ - pE\]
C. \[pE\]
D. \[2pE\]
Answer
301.8k+ views
Hint:A dipole is the combination of two charges of equal in magnitude and opposite in nature. The work done by the external agency is stored as the potential energy of the system. So by finding the change in electric potential energy of the dipole, we can find the work done by the external agency.
Formula used:
\[U = - pE\cos \theta \]
where U is the electric potential energy stored in the dipole system, p is the dipole moment, E is the electric field and \[\theta \] is the angle between the dipole moment vector and the electric field vector.
\[W = \Delta U\]
where W is the work done by the external agency and \[\Delta U\] is the change in potential energy.
Complete step by step solution:

Image: Dipole in electric field
Initially, the dipole moment is parallel to the external electric field vector. When two vectors are parallel to each other, then the angle between the vectors is 0°. So, the initial potential energy of the dipole system is,
\[{U_i} = - pE\cos {\theta _i}\]
\[\Rightarrow {U_i} = - pE\cos 0^\circ \]
\[\Rightarrow {U_i} = - pE\]
Finally, the dipole moment is anti-parallel to the external electric field vector. When two vectors are antiparallel to each other, then the angle between the vectors is 180°.
So, the final potential energy of the dipole system is,
\[{U_f} = - pE\cos {\theta _f}\]
\[\Rightarrow {U_f} = - pE\cos 180^\circ \]
\[\Rightarrow {U_i} = pE\]
The work done by the external agency is the change in electric potential energy of the dipole system.
So, the work done can be calculated as,
\[W = \Delta U\]
\[\Rightarrow W = {U_f} - {U_i}\]
\[\Rightarrow W = pE - \left( { - pE} \right)\]
\[\therefore W = 2pE\]
Hence, the work done by the external agency to rotate the given dipole from the initial position of being parallel to the electric field to the final position of being anti-parallel to the electric field.
Therefore, the correct option is D.
Note: The work done by the restoring force is negative to the change in potential energy of the system and the work done by the external force is equal to the change in potential energy of the system.
Formula used:
\[U = - pE\cos \theta \]
where U is the electric potential energy stored in the dipole system, p is the dipole moment, E is the electric field and \[\theta \] is the angle between the dipole moment vector and the electric field vector.
\[W = \Delta U\]
where W is the work done by the external agency and \[\Delta U\] is the change in potential energy.
Complete step by step solution:

Image: Dipole in electric field
Initially, the dipole moment is parallel to the external electric field vector. When two vectors are parallel to each other, then the angle between the vectors is 0°. So, the initial potential energy of the dipole system is,
\[{U_i} = - pE\cos {\theta _i}\]
\[\Rightarrow {U_i} = - pE\cos 0^\circ \]
\[\Rightarrow {U_i} = - pE\]
Finally, the dipole moment is anti-parallel to the external electric field vector. When two vectors are antiparallel to each other, then the angle between the vectors is 180°.
So, the final potential energy of the dipole system is,
\[{U_f} = - pE\cos {\theta _f}\]
\[\Rightarrow {U_f} = - pE\cos 180^\circ \]
\[\Rightarrow {U_i} = pE\]
The work done by the external agency is the change in electric potential energy of the dipole system.
So, the work done can be calculated as,
\[W = \Delta U\]
\[\Rightarrow W = {U_f} - {U_i}\]
\[\Rightarrow W = pE - \left( { - pE} \right)\]
\[\therefore W = 2pE\]
Hence, the work done by the external agency to rotate the given dipole from the initial position of being parallel to the electric field to the final position of being anti-parallel to the electric field.
Therefore, the correct option is D.
Note: The work done by the restoring force is negative to the change in potential energy of the system and the work done by the external force is equal to the change in potential energy of the system.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
Electron Gain Enthalpy and Electron Affinity Explained

Understanding Uniform Acceleration in Physics

Effective Nuclear Charge for JEE

The shortest range of the fundamental force is associated class 12 physics JEE_Main

Understanding Average and RMS Value in Electrical Circuits

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Class 12 CBSE Physics Sample Paper - Set 7 Preparation PDF Download (Login Required)

Understanding Inertial and Non-Inertial Frames of Reference

Why does capacitor block DC and allow AC class 12 physics JEE_Main

Understanding How a Current Loop Acts as a Magnetic Dipole

Units and Measurements Mock Test for JEE Main 2026-27 Preparation

