# Understanding the energy of a dipole in a uniform electric field

• offscene
In summary, Griffith's E&M problem 4.7 asks to calculate the energy of a dipole in a uniform electric field and I ended up getting a different answer than the one given. I thought that calculating the energy/work done to construct the dipole is the same as dragging two point charges where one is d apart from the other (please correct me if I am wrong) under the effect of a constant electric field. This should be done by just being able to drag the first with just the constant E-field acting on it and then dragging in the second charge (opposite sign) which has both the contribution from the constant E-field as well as the other charge that has already been placed. However, the
offscene
Homework Statement
Griffith's E&M, problem 4.7: Show that the energy of an ideal dipole p in an electric field E is given by $$U=-p \cdot E$$
Relevant Equations
$$W=qV$$
$$V(r) = -\int_{\textrm{reference (take as infinity)}}^{r} E \cdot dl$$
Griffith's E&M problem 4.7 asks to calculate the energy of a dipole in a uniform electric field and I ended up getting a different answer than the one given. I thought that calculating the energy/work done to construct the dipole is the same as dragging two point charges where one is d apart from the other (please correct me if I am wrong) under the effect of a constant electric field. This should be done by just being able to drag the first with just the constant E-field acting on it and then dragging in the second charge (opposite sign) which has both the contribution from the constant E-field as well as the other charge that has already been placed. However, the answer does not come out to be correct if I include the contribution of the field from the already placed charge and I am confused as to why we can ignore this. (Essentially, I am getting the same answer as stated but with an additional term due to the potential of the other charge itself)

It is simply that the "energy of a dipole in an electric field" only includes the interaction energy, not the internal energy of the dipole. In real life, most dipoles are not built from two elementary charges, so calculating the total energy of a system of charges is more complicated.

offscene and topsquark
Your approach of starting from "scratch", i.e. only charges at infinity to start with, is OK if you do it consistently and you don't lose track of your goal. If you want to be consistent, you should also consider the energy involved in separating the positive from the negative charges in order to create the uniform field in the region of interest. That would be consistent with your approach, but it would be a complicated calculation that is unnecessary. Read on.

Your goal is to find the potential energy of "an ideal dipole in an electric field". Whichever way you derive the expression, it should reduce identically to zero in two cases:
1. There is no electric field, i.e. ##\mathbf{E}=0##. This excludes the energy of assembling the charges to create the external field. You correctly ignored this term.
2. There is no dipole, i.e. ##\mathbf{p}=0##. This excludes the energy of assembling the two charges to create the dipole. You did not ignore this term but you should have as @DrClaude already remarked.
You don't show the potential that you derived. Assuming that you derived it correctly, it should be reducible to the form ##~U=-\mathbf{p}\cdot\mathbf{E}+U_0~## where ##U_0## is a constant energy that does not depend on the orientation of the dipole relative to the electric field. Since the zero of potential energy is arbitrary and only changes in potential energy matter, one simply redefines ##U_0=0## to describe the potential energy of "an ideal dipole in an electric field". The conventional form of this potential energy is zero when the dipole moment is oriented perpendicular to the external electric field.

offscene, DrClaude and topsquark

• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
703
• Introductory Physics Homework Help
Replies
11
Views
174
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
493
• Introductory Physics Homework Help
Replies
25
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
498
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
718