
The potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are respectively
A. \[{90^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\]
B. \[{0^0}\,and\,{\rm{ 9}}{{\rm{0}}^0}\]
C. \[{90^0}\,and\,{\rm{ }}{{\rm{0}}^0}\]
D. \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\]
Answer
161.7k+ views
Hint:The values of maximum and minimum of the potential at a point due to an electric dipole will be determined by the angles like if \[\theta = {0^0},\cos \theta = 1\] then the electric potential will be maximum at the dipole axis and if \[\theta = {180^0},\cos \theta = - 1\] then the electric potential will be minimum at the dipole axis
Formula used:
The potential due to this dipole is given as,
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
Where,
p is dipole moment,
r is distance from dipole,
k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\] and \[{\varepsilon _0}\] is permittivity.
Complete step by step solution:
The potential due to this dipole is:
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
If \[\theta = {0^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {0^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the maximum potential,
\[{V_{\max }} = \dfrac{{kp}}{{{r^2}}}\]
Now If \[\theta = {180^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {{180}^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = - }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the minimum potential,
\[{V_{\min }} = - \dfrac{{kp}}{{{r^2}}}\]
Therefore the potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\].
Hence option D is the correct answer.
Note:The electric potential will be zero at the perpendicular axis of the dipole when \[\theta = {90^0}\] . As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. Electric potential obeys superposition principle due to electric dipole as a whole can be sum of potential due to both the charges positive and negative.
Formula used:
The potential due to this dipole is given as,
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
Where,
p is dipole moment,
r is distance from dipole,
k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\] and \[{\varepsilon _0}\] is permittivity.
Complete step by step solution:
The potential due to this dipole is:
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
If \[\theta = {0^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {0^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the maximum potential,
\[{V_{\max }} = \dfrac{{kp}}{{{r^2}}}\]
Now If \[\theta = {180^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {{180}^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = - }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the minimum potential,
\[{V_{\min }} = - \dfrac{{kp}}{{{r^2}}}\]
Therefore the potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\].
Hence option D is the correct answer.
Note:The electric potential will be zero at the perpendicular axis of the dipole when \[\theta = {90^0}\] . As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. Electric potential obeys superposition principle due to electric dipole as a whole can be sum of potential due to both the charges positive and negative.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
