Question

Intensity of an electric field (E) depends on distance r. In case of dipole, it is related as :
A. \[E \propto \dfrac{1}{r}\]
B. \[E \propto \dfrac{1}{{{r^2}}}\]
C. \[E \propto \dfrac{1}{{{r^3}}}\]
D. \[E \propto \dfrac{1}{{{r^4}}}\]

Answer 1

Hint:The electric dipole is a system of two charges that are separated by a finite distance. The dipole has a property called dipole moment which is equal to the product of the charges and the distance of separation between them. The Intensity of an electric field (E) due to an electric dipole(p) is in two positions: along axial position and equatorial position.

Formula used:
Intensity of electric field due to a dipole at axial is given as:
\[\overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}\]
Intensity of electric field due to a dipole at equatorial is given as:
\[\overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{p}{{{r^3}}}\]
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:
As intensity of electric field due to a dipole at axial is:
\[\overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}\]
As intensity of electric field due to a dipole at equatorial is:
\[\overrightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{p}{{{r^3}}}\]
Now here we have the relationship so that we can determine the relationship between the Intensity of an electric field (E) due to an electric dipole(p). Hence from this Intensity of an electric field is inversely proportional to the third power of r.

Hence option C is the correct answer.

Note: As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and the other is a negative charge. These two charges are separated by a distance. Electric potential obeys the superposition principle due to electric dipole as a whole can be the sum of potential due to both the charges positive and negative.