
If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x:y is ?
Answer
161.7k+ views
Hint: To solve this question we need to divide magnitude of the intensity of electric field on axial line at a distance x and magnitude of the intensity of electric field on equatorial line at a distance y so that we get the ratio between x and y.
Formula used:
Magnitude of the intensity of electric field at a distance x on axial line is given as,
\[{E_a} = k \times \dfrac{{2p}}{{{x^3}}}\]
Where p is dipole moment, r is distance from dipole, k is coulomb’s constant and \[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}\]
\[{\varepsilon _0}\] is permittivity
Magnitude of the intensity of electric field at a distance Y on equatorial line is given as,
\[{E_e} = k \times \dfrac{p}{{{y^3}}}\]
Where p is the dipole moment, \[{E_a}\] is the intensity of electric field on the axial line and \[{E_e}\] is the intensity of electric field on the equatorial line.
Complete step by step solution:
Given magnitude of intensity of electric field at a distance x on axial line is equal to magnitude of intensity of electric field at a distance y on equatorial line. As we know magnitude of the intensity of electric field at a distance x on axial line,
\[{E_a} = k \times \dfrac{{2p}}{{{x^3}}}\]
Magnitude of the intensity of electric field at a distance y on equatorial line,
\[{E_e} = k \times \dfrac{p}{{{y^3}}}\]
Now according to question equate both the terms, we get
\[\dfrac{{{E_a}}}{{{E_e}}} = \dfrac{{k \times \dfrac{{2p}}{{{x^3}}}}}{{k \times \dfrac{p}{{{y^3}}}}}\]
By arranging this
\[\therefore \dfrac{x}{y} = \dfrac{{\sqrt[3]{2}}}{1}\]
Therefore the ratio is, \[x:y = \sqrt[3]{2}:1\].
Note: An axial line is defined as the line joining the two charges which is forming an electric dipole and equatorial line is defined as the line which is perpendicular to the axial line which passes through the electric dipole.
Formula used:
Magnitude of the intensity of electric field at a distance x on axial line is given as,
\[{E_a} = k \times \dfrac{{2p}}{{{x^3}}}\]
Where p is dipole moment, r is distance from dipole, k is coulomb’s constant and \[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}\]
\[{\varepsilon _0}\] is permittivity
Magnitude of the intensity of electric field at a distance Y on equatorial line is given as,
\[{E_e} = k \times \dfrac{p}{{{y^3}}}\]
Where p is the dipole moment, \[{E_a}\] is the intensity of electric field on the axial line and \[{E_e}\] is the intensity of electric field on the equatorial line.
Complete step by step solution:
Given magnitude of intensity of electric field at a distance x on axial line is equal to magnitude of intensity of electric field at a distance y on equatorial line. As we know magnitude of the intensity of electric field at a distance x on axial line,
\[{E_a} = k \times \dfrac{{2p}}{{{x^3}}}\]
Magnitude of the intensity of electric field at a distance y on equatorial line,
\[{E_e} = k \times \dfrac{p}{{{y^3}}}\]
Now according to question equate both the terms, we get
\[\dfrac{{{E_a}}}{{{E_e}}} = \dfrac{{k \times \dfrac{{2p}}{{{x^3}}}}}{{k \times \dfrac{p}{{{y^3}}}}}\]
By arranging this
\[\therefore \dfrac{x}{y} = \dfrac{{\sqrt[3]{2}}}{1}\]
Therefore the ratio is, \[x:y = \sqrt[3]{2}:1\].
Note: An axial line is defined as the line joining the two charges which is forming an electric dipole and equatorial line is defined as the line which is perpendicular to the axial line which passes through the electric dipole.
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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

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

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
