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Question: Force of interaction between two co-axis short electric dipoles whose centers are \(R\) distance apa...

Force of interaction between two co-axis short electric dipoles whose centers are RR distance apart varies as:
A) 1R\dfrac{1}{R}
B) 1R2\dfrac{1}{{{R^2}}}
C) 1R3\dfrac{1}{{{R^3}}}
D) 1R4\dfrac{1}{{{R^4}}}

Explanation

Solution

Hint
We need to use the formula of the potential energy of a dipole on its axis due to an external electric field formed by the other dipole. Then we have to differentiate the potential with respect to distance to find the force between the two dipoles.
Formula used: In this solution we will be using the following formula,
Potential energy of a dipole in an external electric field:U=pEcosθU = - pE\cos \theta , pp is the dipole moment, EE is the external electric field, and θ\theta is the angle between the dipole moment and the electric field
Electric field of a dipole on its axis: E=2kpr3E = \dfrac{{2kp}}{{{r^3}}} where kk is the coulomb’s constant, pp is the dipole moment and rr is the distance of the point from the dipole.

Complete step by step answer
We’ve been given that two dipoles lie coaxially with a distance RR between them. We will start by assuming that the first dipole creates an electric field which will affect the second dipole.
The potential energy of the second dipole p2{p_2} when it is placed in the electric field of the first dipolep1{p_1} which lies at a distance RR can be calculated using the formula,
U=p2Ecosθ\Rightarrow U = - {p_2}E\cos \theta
Since, both the dipoles have the same axis, cos0=1\cos 0^\circ = 1, and we can write the value of the electric field as
E=2kp1R3\Rightarrow E = \dfrac{{2k{p_1}}}{{{R^3}}}
On substituting we obtain,
U=2kp1p2R3\Rightarrow U = - \dfrac{{2k{p_1}{p_2}}}{{{R^3}}}
Now the force between these two dipoles, once we know the potential, can be calculated as:
F=dUdR\Rightarrow F = - \dfrac{{dU}}{{dR}}
On differentiating we get,
F=6kp1p2R4\Rightarrow F = \dfrac{{6k{p_1}{p_2}}}{{{R^4}}}
Hence the force is proportional to 1/R41/{R^4} which corresponds to option (D).

Note
Here we have calculated the potential energy of the second dipole when it is placed in the electric field generated by the first dipole. However, it is inconsequential even if we calculate the potential energy of the first dipole when placed in an electric field generated by the second dipole. We can validate this claim since the force between the two dipoles must be equally experienced by both p1{p_1} and p2{p_2} from Newton’s third law.