Question

Force of interaction between two co-axis short electric dipoles whose center’s are R distance apart varies as
(A) $\dfrac{1}{R}$
(B) $\dfrac{1}{{{R^2}}}$
(C) $\dfrac{1}{{{R^3}}}$
(D) $\dfrac{1}{{{R^4}}}$

Answer 1

Hint: Coulomb’s law states that the electrostatic force between two stationary point charges is directly proportional to multiple of their magnitudes and inversely proportional to the square of the distance between their center.
⟹Coulomb’s force will be repulsive or attractive depending upon whether the charges are like or unlike charges. Further, the force always acts along the line joining the centers of the two charges.
Formula used:
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where,
F=force of attraction between stationary point charges
${q_1}$ and ${q_2}$ stationary point charges
r=distance between charges
k is a constant of proportionality called electrostatic force constant

Complete step by step answer:
Consider two stationary point charges q1 and q2 which are at a distance r from each other in a vacuum.
We know that the magnitude of the electrostatic force between the charges is given by,
$F \propto \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ …………. (1)
Where k is a constant of proportionality called electrostatic force constant.
We know that, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$
From equation (1), Force of attraction can be written as,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ …………… (2)
We know that there is an electric dipole that is separated by a distance R then we are interested to find the force of interaction between the co-axis short electric dipole.
For this, we need to apply coulomb’s law to the given situation.
Then electric dipole moment is given as,
$p = 2a \times q$
By putting the values in electric dipole moment we get,
From this we get,
$p = R \times {q_1}$
$ \Rightarrow {q_1} = \dfrac{p}{R}$
Similarly,$p = R \times {q_2}$
$ \Rightarrow {q_2} = \dfrac{p}{R}$
Now substitute these values in equation (2), we get,
\[
  F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{(\dfrac{p}{R})(\dfrac{p}{R})}}{{{R^2}}} \\
   = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{{(\dfrac{p}{R})}^2}}}{{{R^2}}} \\
 \]
Thus, $F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{p^2}}}{{{R^4}}}$
This is the force of interaction between two co-axis short electric dipoles whose centers are R distance.
From this, we can say that force is inversely proportional to the fourth power of the distance between the electric dipole.
$F\propto \dfrac{1}{{{R^4}}}$

Therefore, the correct option is (D).

Additional information:
A pair of two equal and opposite point charges separated by a small distance is called an electric dipole. The electric dipole consists of two equal and opposite charges (-q and) separated by a small distance 2a. The line joining the charges is called the dipole axis. The length of the dipole (2a) is a vector whose direction is from the –q to +q. the total charge of the dipole is equal to zero. But the electric field due to the electric dipole is not zero. It is because the charges –q and +q are separated by some distance and the resultant field due to these changes is not zero.
The dipole moment is a measure of the strength of the electric dipole. The dipole moment of an electric dipole is a vector whose magnitude is equal to the product of either charge and the length of the dipole. That is, $\overrightarrow p = q(\overrightarrow {2a)} $. The direction of p is along the dipole axis from –q to +q.

Note: The main limitation of this coulomb’s law is it is applicable only for the point charges. Means compared to its charges distance must be very large. And this force depends on the medium in which charges are present.