Calculating Energy of a Dipole

[Observer Two]

I have a dipole such as:

[itex]\rho(\vec{r}) = q \delta(\vec{r} - \vec{a}) - q \delta(\vec{r} + \vec{a})[/itex]

with [itex]\vec{a} = a \vec{e}_x[/itex].

I have to show that the energy in a constant external field [itex]\vec{E}[/itex] is:

[itex]V = - 2 q \vec{a} \vec{E}[/itex]

My calculations so far:

With the formula: [itex]\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \! \rho(\vec{r}') \frac{1}{|\vec{r} - \vec{r}'|} \, d^3r' [/itex]

I have calculated for the potential:

[itex]\phi(\vec{r}) = \frac{q}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]

The energy [itex]V(\vec{r}) = \phi(\vec{r}) q[/itex]

Thus I get for the energy:

[itex]V(\vec{r}) = \frac{q^2}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]

However I don't understand why that is equal to [itex]-2q \vec{a} \vec{E}[/itex].

Any ideas?

 

[rude man]

The energy referred to is the difference in energy between the dipole's x and y orientation. Depending on the direction of the p and E vectors, this difference can be positive or negative. So you must be careful with your signs.

So look up the formula for the energy of a dipole in an E field. This expression is a dot product so be aware of your signs. Among other things you have to determine the sign (direction) of your dipole as given in the expression for ρ. Hint: ρ is a line charge density and the units of δ(x) are always 1/x.

Alternatively, you can use the formula for torque on a dipole in an E field, which is also a vector expression, then compute the work needed to change the dipole unit vector from the x to the y direction. This energy can again be positive or negative so again be careful with your signs.

P.S. Unless another useless change has occurred in notation since my day, your answer should indicate a dot product. There is no such thing as AB where A and B are vectors. It's either A dot B or A cross B.

But times change, there is no such thing as relativistic mass anymore, and E = mc^2 is no longer correct if m moves. What a shame ...

 

[Observer Two]

Hello rude man!

We've learned the convention that you usually leave out the dot product sign if it's clear from context how it's meant to be. Just like you'd write 2x instead of "2 times x", we write [itex]\vec{A}\vec{B}[/itex] instead of [itex]\vec{A} \cdot \vec{B}[/itex]

I don't think the + or - signs are the problems here. The charge density is given as it is, so the + or - signs automatically follow correctly from integration. You can't do much wrong integrating delta functions I think. The potential should be correct. My problem is to show why [itex]W = - 2 q \vec{a} \cdot \vec{E}[/itex] with [itex]\vec{E}[/itex] being an external field.

The left hand side should be "potential times q". Where do you get the right hand side from though?

 

[rude man]

Hello rude man!

The left hand side should be "potential times q". Where do you get the right hand side from though?

I guess I don't think of potentials in this case. I guess you can but I don't. I seem to remember a paper where the approach is in that direction. Do a google and I think you can find it. It was by a Princeton co-author, a pdf file.

Reason I don't understand the idea of potentials is that the dipole already has its own energy: -kq^2/(2a)^2, without the introduction of an external E field. But this term with k in it does not appear in the answer. You'd think adding the E field would be a sort of superposition of the dipole's intrinsic potential energy plus a term with E in it. But it isn't. So I view it as the energy associated with rotation, which does provide the right answser.

If you do decide to go it my way:
Did you find the equation for the energy of a dipole in an E field? Or at least the expression for torque on a dipole in an E field? That is how I would proceed.

BTW I still think you need to know the direction of the dipole vector in relation to the direction of the E vector. If you got it backwards you'd get the wrong sign in your answer.

Happy figuring!

 

[HalfLostAndSearching]

I can provide a response to your calculations and question. First, let's clarify some terms and concepts. A dipole is a pair of equal and opposite charges separated by a distance, represented by the symbol \vec{p}. The magnitude of the dipole moment is given by p = qa, where q is the magnitude of the charge and a is the distance between the charges. The direction of the dipole moment is from the negative charge to the positive charge.

In your given equation, \rho(\vec{r}) represents the charge density, with the first term representing the positive charge at \vec{a} and the second term representing the negative charge at -\vec{a}. The potential, \phi(\vec{r}), is the work required to move a unit charge from infinity to a point \vec{r}. The energy, V(\vec{r}), is the potential multiplied by the charge.

Your calculations for the potential and energy are correct. However, to understand why V(\vec{r}) is equal to -2q \vec{a} \vec{E}, we need to consider the definition of an electric dipole moment, \vec{p} = q \vec{a}. This means that the dipole moment is equal to the charge multiplied by the distance between the charges.

Now, let's consider the energy in a constant external field \vec{E}. The energy of a dipole in an external field is given by U = -\vec{p} \cdot \vec{E}. This equation tells us that the energy of a dipole in an external field is equal to the negative dot product of the dipole moment and the electric field. In our case, \vec{p} = q \vec{a} and \vec{E} = E \vec{e}_x. Thus, U = -q \vec{a} \cdot E \vec{e}_x = -q \vec{a} \cdot \vec{E}. Since the dot product of two vectors is equal to the product of their magnitudes multiplied by the cosine of the angle between them, we can rewrite this as U = -qaE\cos{\theta}. In our case, the angle between \vec{p} and \vec{E} is 0 degrees, so \cos{\theta} = 1. This means that U = -qaE.

Finally