Electromagnetism, force between dipole and grounded plane

[Antti]

Homework Statement

An electric dipole is located at a certain distance from a grounded plane. What force does the dipole exert on the plane?

(The answer is to be expressed as an equation. No data were given, only the above text)

Homework Equations

I am not sure about this, but the course is about "classical" electromagnetism. I would suspect that coulombs law should be used since the question is about the force between charged particles/objects.

$$F = \frac{Q_{1}Q_{2}}{4 \pi \epsilon_{0} r^{2}}$$

The Attempt at a Solution

At first I just thought that the grounded plane would have no net charge and thus the dipole and plane could not affect each other. I now know that this is isn't true but I'm not sure why. A coursemate told me that the dipole should be treated as two point charges and that they would have equal but opposite charges (mirrored) in the grounded plane. Unfortunately I didn't get the chance to ask him further questions.

Thankful for help

 

[Roberto Bramb]

If M1=Qd is the dipole moment (Q :charge, d:separation as a vector) the potential field is
V(r)=1/(4*pi*e) M1.grad(1/r)
Having a ground plane, this can be eliminated if you consider an image dipole (symmetrically placed under the ground plane). of moment M2=-M1.
Then the interaction energy will be of the order
W=M1*M2/(4 pi eps r^3)*(angle factor)
where r is the distance between the two dipoles.
The radial force will be
F=-dW/dr=3m1m2/(4 pi eps r^4)*(angle factor)
For the angle factor look at classic SMYTHE-Static&DynamicElectricity- McGraw 1968, p.7

 

[Moetasim]

I would approach this problem by considering the fundamental principles of electromagnetism. The force between two charged particles is given by Coulomb's law, as you have correctly stated in your equations. However, in this case, we are dealing with a dipole and a grounded plane, which introduces some additional complexities.

Firstly, we need to consider the orientation of the dipole with respect to the plane. If the dipole is perpendicular to the plane, then the force exerted on the plane will be zero. This is because the dipole's electric field lines will be parallel to the plane, and thus will not interact with it.

However, if the dipole is parallel to the plane, then we need to consider the individual charges that make up the dipole. As your coursemate mentioned, the dipole can be treated as two point charges with equal but opposite charges. These charges will induce an opposite charge on the grounded plane, creating a dipole-dipole interaction.

The force between two dipoles can be calculated using the equation:

F = \frac{1}{4 \pi \epsilon_{0}} \frac{p_{1}p_{2}}{r^{3}} (1 - 3 \cos^{2} \theta)

Where p1 and p2 are the dipole moments of the two dipoles, r is the distance between them, and θ is the angle between the dipoles.

In this case, we can simplify the equation by assuming that the dipole moment of the grounded plane is much smaller than that of the original dipole. This brings us back to Coulomb's law, where the force between the dipole and the grounded plane can be calculated using the equation you have provided:

F = \frac{Q_{1}Q_{2}}{4 \pi \epsilon_{0} r^{2}}

Where Q1 and Q2 are the charges of the two dipoles.

In summary, the force between a dipole and a grounded plane can be calculated using Coulomb's law, taking into account the orientation and relative sizes of the dipoles. I hope this explanation helps you understand the concept better.