Thank you all ahead of time for your guidance, it is much appreciated.

Currently I'm solving some problems out of Purcell's Electricity and Magnetism. In problem 1.5, the reader is asked to consider a system of four charges of equal magnitude q forming a square quadrupole with side length $2a$, with another charge $-Q$ fixed at the center (origin) of the quadrupole. The goal of the problem is to determine whether the charges are in a stable or unstable equilibrium by finding the change in potential energy $U$ in response to some displacement of the central charge $-Q$ to a point $(x, y)$. In this case, a decrease in energy would indicate an unstable equilibrium.

The distance between the corner charges and the central charge is given by \begin{equation} \sqrt{(\pm a - x)^2 + (\pm a - y)^2} \end{equation}

and each corner charge is separated from the two charges along the sides of the quadrupole by a distance $2a$ and from the charge across the diagonal by $2a\sqrt{2}$. The potential energy of this system of charges is given by \begin{equation} U = \frac{1}{2}\sum_{i\ =\ 1}^N\sum_{j\ \neq \ i }k\frac{q_iq_j}{r_{ij}} \end{equation} which becomes \begin{equation} U(x, y) = kq\left(\frac{2q}{a}+\frac{q}{a\sqrt{2}} -\frac{Q}{a}\left(2\sqrt2\ +\ \frac{x^2 + y^2}{2\sqrt{2}a^2} \right) \right ) \end{equation} Yet in the solution to the problem, Purcell gives only the last term which accounts for the potential energy between the corner charges $q$ and the central charge $-Q$: \begin{equation}U(x, y) = -k\frac{Qq}{a}\left(2\sqrt2\ +\ \frac{x^2 + y^2}{2\sqrt{2}a^2} \right)\end{equation}

My confusion is this: why is it not necessary to account for the potential energy between each of the corner charges in the term for the total potential energy of the system? As far as I can tell, both answers illustrate that there should be a decrease in the potential energy $U$ with any displacement in $x$ and $y$, which indicates that the equilibrium is unstable. But why does Purcell not account for the energy required to assemble the quadrupole?

  • $\begingroup$ You're throwing away an overall constant which does not affect the result. $\endgroup$ – ZeroTheHero Mar 5 '17 at 19:59
  • $\begingroup$ @ZeroTheHero Thank you, that brings some clarity to the issue. So does that mean that if the self-energy of a system (the energy required to assemble the original system) is constant, it is unnecessary to account for the self energy? $\endgroup$ – nguzman Mar 5 '17 at 20:18
  • $\begingroup$ No. I'm saying if you are interested in the stability properties of your system then constant terms don't matter. $\endgroup$ – ZeroTheHero Mar 5 '17 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.