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Question: The force of interaction of two dipoles, if the two dipoles are parallel to each other and placed at...

The force of interaction of two dipoles, if the two dipoles are parallel to each other and placed at distance x apart.

1. 3p1p24πεx4\dfrac{{3{p_1}{p_2}}}{{4\pi {\varepsilon _ \circ }{x^4}}}
2. p1p24πεx4\dfrac{{{p_1}{p_2}}}{{4\pi {\varepsilon _ \circ }{x^4}}}
3. p1p24πεx4\dfrac{{{p_1}{p_2}}}{{4\pi {\varepsilon _ \circ }{x^4}}}
4. p1p23πεx4\dfrac{{{p_1}{p_2}}}{{3\pi {\varepsilon _ \circ }{x^4}}}

Explanation

Solution

First, we will need to find the electrostatic field of dipole p2{p_2} at p1{p_1} . Then we will find the potential energy of two dipoles. In the final step we will differentiate the potential energy to get the Force of interaction between two dipoles.

Complete step-by-step Solution
A dipole is separation of two opposite charges and it is quantified by electric dipole moment and is denoted by p.
As we know electric field of dipole along perpendicular bisector of the axis,
E=p4πεr3\overrightarrow E = - \dfrac{{\overrightarrow p }}{{4\pi {\varepsilon _ \circ }{r^3}}} , where r= distance
ε{\varepsilon _ \circ } = permittivity of free space
E21{E_{21}} is the field due to dipole p1{p_1} at dipole p2{p_2}
E21=p14πεx3{E_{21}} = \dfrac{{{p_1}}}{{4\pi {\varepsilon _ \circ }{x^3}}}
Potential energy of dipole system
U=p2.E21U = - \overrightarrow {{p_2}} .\overrightarrow {{E_{21}}}
U=p2p14πεx3cos(π)U = - {p_2}\dfrac{{{p_1}}}{{4\pi {\varepsilon _ \circ }{x^3}}}\cos (\pi )
Angle between the dipole and electric field is 180 degrees.
U=p1p24πεx3U = \dfrac{{{p_1}{p_2}}}{{4\pi {\varepsilon _ \circ }{x^3}}}
Now, to find the force
F=dUdx=34πεp1p2x4F = - \dfrac{{dU}}{{dx}} = \dfrac{3}{{4\pi {\varepsilon _ \circ }}}\dfrac{{{p_1}{p_2}}}{{{x^4}}}
F is positive, so it is a repulsive force.
Option (1) 3p1p24πεx4\dfrac{{3{p_1}{p_2}}}{{4\pi {\varepsilon _ \circ }{x^4}}}

Additional Information
Electric field due to dipole at a general point
E=14πεpr33cos2θ+1E = \dfrac{1}{{4\pi {\varepsilon _ \circ }}}\dfrac{p}{{{r^3}}}\sqrt {3{{\cos }^2}\theta + 1} , θ\theta =angle between the distance vector and dipole.
Potential due to dipole at a general point
V=pcosθ4πεr2V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _ \circ }{r^2}}}

Note
1. You need to keep in mind the direction of the electric field and dipole.
2. While using the formula of potential energy of dipole, you need to find the angle between field and dipole otherwise you will get the wrong force direction.
3. While finding electric fields, Approximation is made that the length of the dipole is negligible as compared to the distance of the point from the dipole.