Solutions to Laplace Equation - Electricity and Magnetism I - Notes | PHY 481, Study notes of Physics

Download Solutions to Laplace Equation - Electricity and Magnetism I - Notes | PHY 481 and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 14 End of Chapter 4 and Start of Chapter 5 of PS Solutions to Laplace’s equations By now we have gone through several methods for find the electric field and electrostatic potential of charge distributions. The basic tools where Gauss’s law and superposition (integration e.g. for rods, discs etc.). In the last few lectures we have extended the superposition method to the case of metals where we found the induced charge on various metal shapes using the method of images. Now we start using the differential equation method, ie. Poisson’s equation ∇2V = −ρ/ǫ. Actually in many cases we are interested in the electric field and electrostatic potential in places where there is no charge density, so we need to solve the simpler equation ∇2V = 0, which is Laplace’s equation. First we solve a simple problem using this approach. A. Conducting sphere or cylinder in a constant electric field Now consider the case of a grounded conducting sphere placed in an electric field. In this case it is not clear how to use the method of images as there are no charges defined in the problem. Nevertheless we can use what we know about the solutions to Laplace’s equation to solve the problem. We know the form of the multipole expansion and taking the monopole and dipole terms we have V (r, θ) = A/r + B + Ccosθ/r2 where the angle θ is with respect the the electric field direction. In addition a constant electric field, E0, in the k̂ direction leads to a potential −E0z = −E0rcosθ. Collecting these terms by superposition we argue that the potential can have the general form, V (r, θ) = A r + B + C cosθ r2 + Drcosθ (1) We know that each term in this expression satisfies Laplaces equation, so the sum does too - that is superposition again. Note that this is not the general form of the solution in spherical co-ordinates, but it will turn out to be sufficient for the problem of a conducting sphere in an electric field. We don’t expect the first two terms to contribute as the first corresponds to a net charge on the sphere and the second to a finite potential. By symmetry we expect the first to be zero and the second is zero as the sphere is grounded. To ensure that the solution is correct as r → ∞, we need D = −E0. Now notice that the last two terms have 1 the same angular dependence and cancel on the surface of the sphere if we choose C = a3E0, therefore the solution is, V (r, θ) = −E0cosθ[r − a3 r2 ] (2) It is then easy to show that Er = E0(1 + 2a3 r3 )cosθ; Eθ = −E0(1 − a3 r3 )sinθ (3) Notice that Eθ(a, θ) = 0 as it must be to ensure that the electric field at the surface of the conductor is perpendicular to the surface. The charge density at the surface is σ(θ) = 3ǫ0E0cosθ. Integrating over the sphere surface gives the net charge on the sphere, Q = ∫ π 0 2πa2sinθ3ǫ0E0cosθdθ = 0 (4) which shows that no charge is transferred from ground to the sphere. In that case this result also applies when the sphere is isolated. However if we integrate over the top hemisphere, Q = ∫ π/2 0 2πa2sinθ3ǫ0E0cosθdθ = 3πa 2ǫ0E0 (5) we see that a positive charge is induced in the top hemisphere. There is an equal amount of negative charge on the lower hemisphere. In a multipole expansion this system has no monopole term, but it has a dipole term. The dipole strength is proportional to the applied electric field. If the sphere is charged, with an additional charge Q0, this additional charge is distributed uniformly on the sphere surface and its effect on the potential is found by superposition (see e.g. Griffiths problem 3.20), i.e. we add kQ0/r to the potential. If we raise the potential of the sphere to a value V0, this is equivalent to placing a charge Q0 = aV0/k on the sphere. Exactly the same procedure works for conducting cylindrical in a constant applied electric field. In this case, the form of the potential is, V (r, φ) = Alnr + B + C cosφ r + Drcosφ → −E0rcosφ + E0 a2cosφ r , (6) by following the same reasoning as for the spherical case above. The electric field compo- nents are Er = E0cosφ(1+a 2/r2); Eθ = −E0sinφ(1−a 2/r2) and the surface charge density is 2ǫ0E0cosφ. Integration shows that there is no net charge transfer to the conducting cylinder so this solution also applies for an isolated conduting cylinder in a constant applied field. Addition of a charge to an isolated cylinder leads to a uniform charge density on the 2