Electric Potential Energy and Multipole Expansion in Electrostatics - Prof. Phillip Duxbur, Study notes of Physics

Download Electric Potential Energy and Multipole Expansion in Electrostatics - Prof. Phillip Duxbur and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 9 Sections 3.7-3.8 of PS A. The electric potential energy of a charge distribution Potential energy The potential energy of a charge q at position ~r is U = qV (~r). In the case of a continuum this becomes ∫ ρ(~r)V (~r)d~r. However in using this continuum equation it is assumed that the potential is unaltered by the added charge ∫ ρ(~r)d~r. Potential energy stored in a charge distribution A different question is: What is the potential energy of a distribution of charges, that is, what is the potential energy stored in a distribution of charges. In this case we muts take into account the way in which the electrostatic potential changes as charge is added to the system. The potential energy stored in a distribution of charges is equal to the work done in setting up the distribution of charges, provided there is no dissipation and no kinetic energy is generated. To set up a distribution of charges Qi at positions ~ri, we need to bring each of the charges in from infinity and place it at its allocated position. The work required to place the first charge is zero (no other charges are there yet). The work required to place the second charge is Q2V21, where V21 = kQ1/r21 is the electric potential at position ~r2 due to charge Q1. Note that r21 = r12 = |~r2 − ~r1|. The work required to place charge 3 at its position is equal to Q3V31 + Q3V32, and so on, once all of the n charges are in position, we have, Un = 1 2 n,n∑ i6=j kQiQj rij = n,n∑ i<j kQiQj rij (1) In these expressions each pair interaction is counted once and the total potential energy is the sum of the potential energies of all pairs. Note: in writing this energy we have ignored the self-energy of each charge. The self-energy is n ∗ Eself and is the same regardless of where the charges are places. Un is the interaction energy between the charges. Taking the continuum limit the electric potential due to a charge distribution is, U = 1 2 ∫ ρ(~r)ρ(~r′)d~rd~r′ 4πǫ0|~r − ~r′| = 1 2 ∫ ρ(~r)V (~r)d~r (2) Note that this is 1/2 the value which would be true if the potential were fixed and when the charge ∫ ρ(~r) as added. This factor of two is thus quite fundamental - it is also a source of considerable confusion. 1 The energy stored in the electric field Using the relations, ρV = −ǫ0(∇ 2V )V (3) Using the vector identity ~∇· (V ~E) = ~∇V · ~E +V ~∇· ~E and E = −~∇V , this may be rewritten as. −ǫ0(∇ 2V )V = −ǫ0~∇ · (V ~∇V ) + ǫ0(~∇V ) 2 (4) Using Gauss’s theorem the first term on the RHS becomes ∮ V ~∇V · d ~A which goes to zero at r → infinity. The only surviving term is the last term on the RHS, so that the energy density in the electric field may then be written as, u(~r) = 1 2 ǫ0 ~E 2 (5) where we used the fact that ~E = −~∇V . This is the energy density in the electric field and is the energy required to set up the charge distribution. Problem 3.28 can be solved by integrating this energy density over the volume of interest. Note that if we integrate the field due to an isolated charge we get infinity!. However we are interested in changes in potential energy due to changing the charge configuration. The intinite self energy of each charge is there no matter what the charge arrangement is, so it plays no role in the physics of the problem. However it is important in trying to formulate a quantum version of EM. B. The multipole expansion The multipole expansion is a systematic perturbation theory of the general expressions for the potential due to a charge distribution, V (~r) = ∑ i kqi |~r − ~ri| or ∫ kρ(~r′)d~r′ |~r − ~r′| . (6) The perturbation expansion requires that the locations of the charges ~ri are all significantly less than the position at which we plan to find the electrostatic potential and the electric field, i.e. ~r >> ~ri. In the case of the continuum integral, the charge density ρ(~r ′) must have a maximum extent which is much less than ~r. The result of the expansion for the potential is, V (~r) = A r + B r2 + C r3 + ..... (7) 2