
An electric dipole is placed in an electric field generated by a point charge.
A. The net electric force on the dipole must be zero
B. The net electric force on the dipole may be zero.
C. The torque on the dipole due to the field must be zero.
D. The torque on the dipole due to the field may be zero.
Answer
216.3k+ views
Hint:As we know, the same charges repel each other and opposite charges attract each other. As we know that for a positive charge( +q) the direction of the electric force is in the direction of the electric field and for a negative charge (–q) the direction of the electric force is in the opposite direction of the electric field.
Formula used:
Force on the dipole is given as,
\[F = (p \cdot \nabla )E\]
Where p is the dipole moment and E is the electric field.
Torque is given by,
\[\tau = \overrightarrow p \times \overrightarrow E \]
Complete step by step solution:
As the electric field is non-uniform. So the magnitude of forces is not equal as it depends on the strength of the electric field. Therefore net force is not equal to zero. Force on the dipole is:
\[F = (p \cdot \nabla )E\]
Here the electric field is not always equal to zero i.e. \[E \ne 0\]
As torque is given by,
\[\tau = \overrightarrow p \times \overrightarrow E = pE\sin \theta \]
If the dipole moment and electric field are parallel then \[\theta = {0^0}\]. Now we have,
\[\begin{array}{l}\tau = pE\sin \theta \\ \Rightarrow \tau {\rm{ = pEsin}}{{\rm{0}}^0}\\ \Rightarrow \tau {\rm{ = 0}}\end{array}\]
Here the torque is zero. Hence if the dipole moment and electric field are parallel to each other then the torque may be zero. Therefore we concluded that if an electric dipole is placed in an electric field generated by a point charge then the torque on the dipole due to the field may be zero.
Hence option D is the correct answer.
Note: The dipole moment is a vector quantity with a magnitude equal to the product of charge and the distance between them and a direction will be from the negative charge to the positive charge. The torque can be measured as the cross product of dipole moments and the force acting on the dipole.
Formula used:
Force on the dipole is given as,
\[F = (p \cdot \nabla )E\]
Where p is the dipole moment and E is the electric field.
Torque is given by,
\[\tau = \overrightarrow p \times \overrightarrow E \]
Complete step by step solution:
As the electric field is non-uniform. So the magnitude of forces is not equal as it depends on the strength of the electric field. Therefore net force is not equal to zero. Force on the dipole is:
\[F = (p \cdot \nabla )E\]
Here the electric field is not always equal to zero i.e. \[E \ne 0\]
As torque is given by,
\[\tau = \overrightarrow p \times \overrightarrow E = pE\sin \theta \]
If the dipole moment and electric field are parallel then \[\theta = {0^0}\]. Now we have,
\[\begin{array}{l}\tau = pE\sin \theta \\ \Rightarrow \tau {\rm{ = pEsin}}{{\rm{0}}^0}\\ \Rightarrow \tau {\rm{ = 0}}\end{array}\]
Here the torque is zero. Hence if the dipole moment and electric field are parallel to each other then the torque may be zero. Therefore we concluded that if an electric dipole is placed in an electric field generated by a point charge then the torque on the dipole due to the field may be zero.
Hence option D is the correct answer.
Note: The dipole moment is a vector quantity with a magnitude equal to the product of charge and the distance between them and a direction will be from the negative charge to the positive charge. The torque can be measured as the cross product of dipole moments and the force acting on the dipole.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Average and RMS Value in Electrical Circuits

