
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
139.5k+ 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
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

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

A point charge + 20mu C is at a distance 6cm directly class 12 physics JEE_Main

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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 Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

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