Question

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 1

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.