Dipole and a conducting grounded plane

In summary, the equation for the force is wrong, the potential energy is correct, and the procedure for finding the potential energy is not explained.
  • #1
gnegnegne
12
1

Homework Statement


A dipole ##\textbf{p}=p\mathbf{\hat{z}}## (oriented along the z axis for example) is parallel to a conducting grounded plane (xz plane) and is placed at a distance ##d## from it. I have to find the interaction energy and the force between the dipole and the plane.

Homework Equations


Method of image charges,
Electric field of a dipole: [itex]\textbf{E}=\frac{1}{4\pi \epsilon_0}\Big[\frac{3(\textbf{(pr)r}}{r^5} - \frac{\textbf{p}}{r^3} \Big][/itex]
Potential energy of a dipole in an electric field: [itex]U=-\textbf{pE}[/itex]
Force on a dipole in an electric field: [itex]\textbf{F}=\nabla\textbf{(pE)}[/itex]

The Attempt at a Solution


The image charges is a dipole antiparallel to the original one and on the same axis. The electric field along the y-axis generated by the image dipole is [itex]\mathbf{E}=\frac{p}{4\pi\epsilon_0(2y)^3}\mathbf{\hat{z}}=\frac{p}{32\pi\epsilon_0y^3}\mathbf{\hat{z}}[/itex]. The force acting on the original dipole is therefore [itex]\mathbf{F}=-\frac{3p^2}{32\pi\epsilon_0y^3}\mathbf{\hat{y}}[/itex] evaluated at ##y=d##. To calculate the potential energy I can use the other equation [itex]U=-\textbf{pE}[/itex] or I could also calculate the work done by the force from ##y=\infty## to ##y=d##. The result is the same, which is [itex]U=-\frac{p^2}{32\pi\epsilon_0y^3}[/itex]. But is this the correct potential energy? In the example of the method of image charges with a single charge and a conducting plane it is often shown that if you calculate the potential energy [itex]U=-\frac{q^2}{4\pi\epsilon_02y}[/itex] you get twice the correct energy because you are including also the energy of the image charge. Should I consider half the potential energy I calculated? How are the two examples different?
 
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  • #2
gnegnegne said:

Homework Statement


A dipole ##\textbf{p}=p\mathbf{\hat{z}}## (oriented along the z axis for example) is parallel to a conducting grounded plane (xz plane) and is placed at a distance ##d## from it. I have to find the interaction energy and the force between the dipole and the plane.

Homework Equations


Method of image charges,
Electric field of a dipole: [itex]\textbf{E}=\frac{1}{4\pi \epsilon_0}\Big[\frac{3(\textbf{(pr)r}}{r^5} - \frac{\textbf{p}}{r^3} \Big][/itex]
Potential energy of a dipole in an electric field: [itex]U=-\textbf{pE}[/itex]
Force on a dipole in an electric field: [itex]\textbf{F}=\nabla\textbf{(pE)}[/itex]

The Attempt at a Solution


The image charges is a dipole antiparallel to the original one and on the same axis. The electric field along the y-axis generated by the image dipole is [itex]\mathbf{E}=\frac{p}{4\pi\epsilon_0(2y)^3}\mathbf{\hat{z}}=\frac{p}{32\pi\epsilon_0y^3}\mathbf{\hat{z}}[/itex]. The force acting on the original dipole is therefore [itex]\mathbf{F}=-\frac{3p^2}{32\pi\epsilon_0y^3}\mathbf{\hat{y}}[/itex] evaluated at ##y=d##. To calculate the potential energy I can use the other equation [itex]U=-\textbf{pE}[/itex] or I could also calculate the work done by the force from ##y=\infty## to ##y=d##. The result is the same, which is [itex]U=-\frac{p^2}{32\pi\epsilon_0y^3}[/itex]. But is this the correct potential energy? In the example of the method of image charges with a single charge and a conducting plane it is often shown that if you calculate the potential energy [itex]U=-\frac{q^2}{4\pi\epsilon_02y}[/itex] you get twice the correct energy because you are including also the energy of the image charge. Should I consider half the potential energy I calculated? How are the two examples different?
I do not understand your procedure after finding the force.
Your ##U=\vec p.\vec E## equation is for the rotational PE of a dipole, not for moving it within the field. (Within a uniform field, keeping the dipole in the same orientation, no work is done.)
Integrating F.dy looks right, but I don't understand how it went from 3/(32y3) to 1/(32y3) (instead of to 3/(64y2)).

Wrt PE of point charge, I'd need to see an example of the procedure that leads to the wrong answer.
 
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  • #3
Thank you for the answer!
The equation for the force is wrong, I forgot to edit the 3 at the exponent, it's a 4 (you get it when you derive the electric field). So [itex]\mathbf{F}=-\frac{3p^2}{32\pi\epsilon_0y^4}\mathbf{\hat{y}}[/itex]. The other equations should be right. About the potential energy: I'm pretty sure that the formula I wrote is the potential energy of a dipole in a generic field, see for example http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html. For the example of the point charge, instead, see here http://farside.ph.utexas.edu/teaching/em/lectures/node64.html.
 
  • #4
gnegnegne said:
About the potential energy: I'm pretty sure that the formula I wrote is the potential energy of a dipole in a generic field, see for example http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html.
As I posted, that formula is for the PE stored in a dipole that is rotated at some angle within a uniform field (as opposed to aligned with it). That is what the page you linked is all about. It has nothing to do with the energy that would be required to move a dipole near a conducting plate to infinity.
gnegnegne said:
For the example of the point charge, instead, see here http://farside.ph.utexas.edu/teaching/em/lectures/node64.html.
I couldn't find in there anything about a procedure that gives the wrong answer. Can you mention a text string to search for?
 
  • #5
It's a scalar product, of course the energy depends on the angle between the dipole and the electric field, but it is not necessary to have an uniform electric field. See here for example https://physics.stackexchange.com/q...-energy-of-a-dipole-in-an-external-electric-f. It is the general expression and in this case the electric field depends on the position on the y-axis.
For the other thing check "What, finally, is the potential energy of the system. For the analogue problem this is just...".. It calculates the energy of the system of the two charges and then the one of the original system.
 
  • #6
gnegnegne said:
It's a scalar product, of course the energy depends on the angle between the dipole and the electric field, but it is not necessary to have an uniform electric field. See here for example https://physics.stackexchange.com/q...-energy-of-a-dipole-in-an-external-electric-f. It is the general expression and in this case the electric field depends on the position on the y-axis.
I do not know how I am going to convince you that you are confusing two entirely different energies. The links you post for this equation are all to do with energy associated with the torque on the dipole. The energy is released by allowing the dipole to rotate to be be antiparallel with the field.
In the set-up in this thread there will be no torque. The induced field (at the dipole) will be antiparallel to the dipole axis. The energy referred to in this question, as I understand it, is the energy to take the dipole away to infinity.
 
  • #7
gnegnegne said:
For the other thing check "What, finally, is the potential energy of the system. For the analogue problem this is just...".. It calculates the energy of the system of the two charges and then the one of the original system.
Ok, I understand what is going on there. The point is that as you move the point charge away from the plate by some dx the image charge also moves dx away for no extra work done. But integrating the force (as a function of the full distance, 2x) gives the right answer. You would get the wrong answer (double) if you replace the plate by a real second dipole and find the energy of that arrangement.
 
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  • #8
Sorry, I don't want to be arrogant, but I just don't understand what I'm missing.
Suppose you have a dipole with a charge ##-q## at ##\mathbf{r}## and a charge ##q## at ##\mathbf{r}+\mathbf{d}##. In the region there is a potential ##V(\mathbf{r})## (generated by something else). The total potential energy of the dipole is the sum of the energies of the two charges: ##U=-qV(\mathbf{r})+qV(\mathbf{r}+\mathbf{d})##. Expanding it you get ##U=q[V(\mathbf{r})+\mathbf{d}\nabla V(\mathbf{r})-V(\mathbf{r})]=\mathbf{p}\cdot \nabla V=-\mathbf{p}\cdot \mathbf{E}=-pE\cos \theta##. This is the derivation we have seen in class. If the field is uniform the potential energy depends only on the angle and is minimum when the dipole is aligned. But in the derivation we didn't consider an uniform field, so the formula is more general. In this particular case the field and the dipole are aligned, but the field isn't uniform. Of course you can get the torque from the energy using the virtual work principle, but the nature of the energy is purely electric. Why do you say they are two entirely different energies? The only thing I used is the work done to bring the charges from infinity (qV).

About the original problem I managed to find the answer. The force is half the one I calculated (I found a similar exercise in a book with the answer): first you get the force between two dipoles placed at a certain distance ##d## and then you substitute ##d=2y##, while I substituted before deriving the field. This is coherent with what was expected: the energy is stored in the fields, but in this case you have to consider just half of the space because (##y>0##) like is written in the links above.

Thank you very much for your help and sorry for my stubborness, but I really don't understand what's wrong in my argument about the energy of a dipole :)
 
  • #9
gnegnegne said:
Thank you very much for your help and sorry for my stubborness, but I really don't understand what's wrong in my argument about the energy of a dipole :)
No, it is I that should apologise to you. I have been objecting to that equation for the wrong reason.
The right reason is effectively the same as the flaw you linked to for the point charge. It is that as the dipole moves, the image moves, so the field is not fixed. The equation assumes that the field is static.

So now the question is, how come it produces the right answer in the dipole case? I'll try to get back to that, but I'm a bit busy today.
 

Related to Dipole and a conducting grounded plane

1. What is a dipole?

A dipole is a type of antenna that consists of two equal and opposite electric charges separated by a small distance. It is commonly used in radio and communication systems.

2. What is a conducting grounded plane?

A conducting grounded plane is a flat surface made of a conductive material, such as metal, that is connected to the ground. It is often used as a reflector or ground plane for antennas.

3. How do a dipole and a conducting grounded plane interact?

When a dipole is placed near a conducting grounded plane, the plane acts as a mirror, reflecting the electromagnetic waves emitted by the dipole. This results in an increase in the strength and directionality of the radiation pattern of the dipole.

4. What are the benefits of using a dipole and a conducting grounded plane together?

Using a dipole and a conducting grounded plane together can improve the efficiency and directionality of the antenna system. It can also help reduce interference and improve signal strength and quality.

5. Are there any limitations to using a dipole and a conducting grounded plane?

One limitation of using a dipole and a conducting grounded plane is that they are most effective at certain frequencies and may not work well for a wide range of frequencies. Additionally, the size and spacing of the dipole and plane must be carefully designed for optimal performance.

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