Can a dipole be modeled as a point charge?

[Electrodude]

Homework Statement

I was solving a question and tried solving it by modelling an electric dipole as a point charge

Relevant Equations

dipole electric field on axis (short dipole, far away) = (2kp)/r^3

I was trying to solve this question when I got this idea: If the electric field due to a dipole on its axis, far from the short dipole is given by (2kp)/r^3, which we can write as (k(2p/r))/r^2 here this is similar to the electric field due to a point charge, but our charge is of a special type which has magnitude equal to 2p/r. This is not general as we have only come up with this using the field on the axis. Our special type of charge varies inversely with the distance from it. So with this in mind, can we model the dipole as a point charge with magnitude 2p/r and calculate the attractive force between it and the linear charge density. Im not getting the answer that way and know it is somehow wrong, in the solution video (from QR) also they use an unfamiliar formula, so i want to know how to solve this and why my approach is wrong.
Thanks in Adv.

 

[kuruman]

The dipole moment is constant and equal to p. Your expression 2p/r is not constant and depends on the distance of the dipole from wherever r is measured. Furthermore, the electric field lines generated by your "dipole" do not look like dipolar field lines. Even though far from the dipole they look like they vary as $r^{-3}$, they are not spherically symmetric. If you come closer there is a dependence on $\theta$ which is the angle between the direction of the dipole and the position vector of your point of observation.

The safe approach is to find an expression for the potential energy $U$ and then find the force from the potential energy.

(Edited for typo)

 

[TSny]

A dipole has zero net charge. So, modeling the dipole as a system with a variable net charge (2q/r) (2p/r) [edited] is not going to work. Try your model for the case where the dipole is in a uniform electric field as a simple example that shows your model doesn’t work.

To derive the answer to the question, model the dipole as two charges $q$ and $-q$ separated by a small distance x, and where the center of the dipole is a distance $r$ from the line charge.

Derive an expression for the net force on the dipole. Then, simplify your answer by assuming that you have an ideal dipole where $x$ goes to zero and $q$ goes to infinity such that the product $qx = p$ remains constant.

[Edit: I see @kuruman posted just ahead of me. ]

 

[Orodruin]

If you come closer there is a dependence on θ which is the angle between the direction of the dipole and the position vector of your point of observation.

This dependence is present also far away from the dipole.

 

[kuruman]

This dependence is present also far away from the dipole.

Yes.

 

[Orodruin]

Yes.

Your post seemed to indicate otherwise …

 

[kuruman]

Your post seemed to indicate otherwise …

Perhaps. I could have argued that the dipole potential is $V=\dfrac{kp\cos\theta}{r^2}$ and derived the electric field from that. I also thought of saying that OP's proposed approximation does not have the appropriate dipolar symmetry about the plane perpendicular to the moment. Eventually I decided to say what I thought would be enough in OP's terms to show why this approximation is invalid without going through details that might be beyond OP's grasp.

 

[Electrodude]

Thanks @kuruman @Orodruin and @TSny for taking the time to engage.
It seems I was not clear in the original post, I am not trying to model the whole of the dipole that way, that I understand is not possible. Instead what I'm trying to do is model the electric dipole's behaviour on the axis alone, so for anyone traversing the axis at a far enough distance, they won't be able to distinguish between how a dipole would behave and our special charge would behave, In terms of the electric field produced.

But I think I got an idea of how this is wrong, I think the model would work for point charges on the axis, but not for distributions that span across a large enough angle from the dipole's point of view.

This is because, as you all have pointed out, is due to the unsymmetric field of a dipole and the symmetric field of our model, as there will be force vectors oblique to the dipole axis for the case in the question I posted (the image), and for those vectors the horizontal components alone add up.
So I seemed to have answered myself, but I just want to know if that reasoning is right.

Thanks in advance.

 

[TSny]

I think the model would work for point charges on the axis, but not for distributions that span across a large enough angle from the dipole's point of view.

Suppose you want to find the force on a dipole $p$ produced by a linear charge distribution with uniform linear charge density $\lambda$ as shown.

The force that the linear charge distribution exerts on the dipole is equal and opposite to the net force that the dipole exerts on the charge distribution. The latter force can be written as $$F = \int E_{\rm dipole}dq = \int_a^{a+L} E_{\rm dipole}\lambda dx = \lambda \int_a^{a+L} \frac{2kp}{x^3}dx$$.

Here, it is necessary to note that the linear charge density lies along the axis of the dipole so that we can use the expression for the field of a dipole along its axis.

If I’m following you, you would like to write the field of the dipole along its axis as $E = k \frac{2p/x}{x^2}$ so that you can model the field as the field of a point charge where the value of the point charge ($2p/x$) is not a constant but depends on the distance $x$. I don’t see any advantage in looking at it this way.

 

[kuruman]

I don’t see any advantage in looking at it this way.

Neither do I, but why should we look at this situation at all? In the original question, the dipole moment is along the $x$-axis and the linear charge distribution is along $z$. One can find the force as you suggest in post #3.

However, it looks like none of the provided answers is correct. The dipole moment points to the right as shown in the figure. If the dipole were drawn as charges +q and -q separated by some distance d, charge -q would be closer to the positive distribution than charge +q. This means that the net force on the dipole is attractive, i.e. the force vector is opposite to the dipole moment vector. The given choices with the dipole oriented as shown are appropriate to a negative linear charge distribution or to a question that asked for the magnitude of the force.

 

[TSny]

Neither do I, but why should we look at this situation at all? In the original question, the dipole moment is along the $x$-axis and the linear charge distribution is along $z$. One can find the force as you suggest in post #3.

Yes. From post #8, @Electrodude realizes that the "point charge model" won't work for the original problem, but the model should work for charge distributions along the axis of the dipole.

However, it looks like none of the provided answers is correct. The dipole moment points to the right as shown in the figure. If the dipole were drawn as charges +q and -q separated by some distance d, charge -q would be closer to the positive distribution than charge +q. This means that the net force on the dipole is attractive, i.e. the force vector is opposite to the dipole moment vector. The given choices with the dipole oriented as shown are appropriate to a negative linear charge distribution or to a question that asked for the magnitude of the force.

I agree. Perhaps the question is asking for just the magnitude of the force.