Question

The distance between the two charges +q and −q of a dipole is r. The intensity at a point on the axial line at a distance x from the center of the dipole is proportional to
A. \[\dfrac{q}{{{x^2}}} \\ \]
B. \[\dfrac{{qr}}{{{x^2}}} \\ \]
C. \[\dfrac{q}{{{x^3}}} \\ \]
D. \[\dfrac{{qr}}{{{x^3}}}\]

Answer 1

Hint:Before we start addressing the problem, we need to know about the electric field intensity due to the dipole. When a unit positive charge placed at a point experiences a force, this force is defined as the electric field intensity. If the intensity is directed away from the charge, then it is due to the influence of a positive charge and if it is directed towards the charge then it is due to a negative charge.

Formula Used:
To find the magnitude of the intensity of the electric field, we have
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{2P}}{{{x^3}}}\]
Where, P is the electric dipole of the dipole moment, x is the distance from the center of a dipole and \[{\varepsilon _0}\] is the permittivity of free space.

Complete step by step solution:

Image: Two charges of a dipole separated by a distance r.

Consider the two charges +q and −q of a dipole which is separated by a distance r. we need to find the intensity at a point on the axial line at a distance x from the center of the dipole. We know the formula to find the magnitude of the intensity of the electric field is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{2P}}{{{x^3}}}\]
Here, P is the dipole moment, \[P = qr\], here q is the charge of a dipole and r is the distance between the two dipoles.

Substituting the value of P in above equation we get,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{2qr}}{{{x^3}}} \\ \]
\[ \therefore E \propto \dfrac{{qr}}{{{x^3}}}\]
Therefore, the intensity at a point on the axial line at a distance x from the center of the dipole is proportional to \[\dfrac{{qr}}{{{x^3}}}\].

Hence, option D is the correct answer.

Note: Suppose if we have two charges, one is positive and one is negative then, the electric dipole gives the direction of the flow of charges. That is, from negative charge toward positive charge.