Solutions to Laplace Equation - Electricity and Magnetism - Lecture Notes, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Solutions to Laplace Equation - Electricity and Magnetism - Lecture Notes and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! 1 PHY481 - Lecture 12: Solutions to Laplace’s equation Griffiths: Chapter 3 Before going to the general formulation of solutions to Laplace’s equations we will go through one more very important problem that can be solved with what we know, namely a conducting sphere (or cylinder) in a uni- form field. We will solve this problem using a superposition of solutions including the dipole potential we found above. Conducting sphere or cylinder in a constant electric field We consider a grounded conducting sphere placed in a uniform 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. The potential due to a uniform electric field ~E = E0ẑ is −E0z or in polar co-ordinates −E0rcosθ. If we combine this with the point charge and dipole potentials, with arbitrary constants, we have, V (r, θ) = A r +B + C cosθ r2 +Drcosθ r > a (1) where a is the radius of the sphere. We know that each term in this expression satisfies Laplace’s 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 term to be zero while the second term is zero because the sphere is grounded. To ensure that the solution is correct as r → ∞, we need D = −E0. Since the conducting sphere is grounded, we must have the boundary conditions that the potential at the surface of the sphere be equal to zero V (r = a, θ) = 0. If we choose C = a3E0, A = B = 0 then we satisfy the boundary conditions on the potential so we suspect the solution is, V (r, θ) = −E0cosθ[r − a3 r2 ] = −E0z + k ~pinduced · r̂ r2 (2) where the induced dipole moment is given by, pinduced = E0a3/k. The induced dipole moment is then proportional to the electric field. The electric field outside the sphere is given by, Er = − ∂V ∂r = E0(1 + 2a3 r3 )cosθ; Eθ = − 1 r ∂V ∂θ = −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. At this point we have satisfied all the boundary conditions so we know the solution is correct and unique. The induced charge density at the surface of the sphere is calculated from σ(θ) = 0Er(r = a) = 30E0cosθ. Integrating over the sphere surface gives the net charge on the sphere, Q = ∫ π 0 2πa2sinθ(30E0cosθ)dθ = 0 (4) which shows that no charge is transferred from ground to the sphere, so this result also applies when the sphere is isolated. Moreover if we integrate over the top hemisphere, Q = ∫ π/2 0 2πa2sinθ(30E0cosθ)dθ = 3πa20E0 (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. Note that the net force on the conducting sphere is zero, as the net force on a dipole in a uniform field is zero. A conducting or dielectric sphere is attracted into an electric field, in order to screen out the field and hence reduce the energy. This is the principle that allows optical tweezers (lasers) to manipulate quantum dots and biomolecules tethered to latex spheres. 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. Exactly the same procedure works for conducting cylindrical in a constant applied electric field, where the solution is, V (r, φ) = −E0cosφ(s+ a2 s ), (6) 2 by following the same reasoning as for the spherical case above. The electric field components are Er = E0cosφ(1 + a2/r2); Eθ = −E0sinφ(1 − a2/r2) and the surface charge density is 20E0cosφ. Integra- tion shows that there is no net charge transfer to the conducting cylinder so this solution also applies for an isolated conducting cylinder in a constant applied field. Addition of a charge to an isolated cylinder leads to a uniform charge density on the surface of the cylinder and an additional term Alnr in the potential. NOTE: this calculation does not allow the cylinder to move. What if we allow the cylinder to rotate? What orientation do you think it will adopt? Laplace’s equation in the three co-ordinate systems Now we are ready to look at more general procedures for solving Laplace’s equation, ∇2V = 0. The Laplacian in any othogonal co-ordinate system is (see Lecture 4), ∇2V = 1 h1h2h3 [ ∂ ∂u1 ( h2h3 h1 ∂V ∂u1 ) + ∂ ∂u2 ( h1h3 h2 ∂V ∂u2 ) + ∂ ∂u3 ( h1h2 h3 ∂V ∂u3 )] (7) For Cartesian co-ordinates we have u1 = x, u2 = y, u3 = z, and h1 = 1, h2 = 1, h3 = 1, so that, ∇2V = ∂ 2V ∂x2 + ∂2V ∂y2 + ∂2V ∂z2 . (8) In cylindrical co-ordinates u1 = s, u2 = φ, u3 = z, and h1 = 1, h2 = s, h3 = 1, so that, ∇2V = 1 s ∂ ∂s (s ∂V ∂s ) + 1 s2 ∂2V ∂φ2 + ∂2V ∂z2 . (9) In spherical polar co-ordinates u1 = r, u2 = θ, u3 = φ, and h1 = 1, h2 = r, h3 = rsinθ, so that, ∇2V = 1 r2 ∂ ∂r (r2 ∂V ∂r ) + 1 r2sinθ ∂ ∂θ (sinθ ∂V ∂θ ) + 1 r2sin2θ ∂2V ∂φ2 . (10) Problems with dependence on only one co-ordinate (i.e., z or s or r) In Cartesian co-ordinates, this corresponds to the potential outside a sheet of charge, or equivalently the potential of a constant electric field −E0z. In fact the general solution is quite trivial, namely, ∂2V ∂z2 = 0 so V (z) = a+ bz (11) where a and b are two constants that can be used to fit the boundary conditions. The solution is thus always linear. In cylindrical co-ordinates, this corresponds to the potential outside a line charge, so we expect a logarithmic solution to be possible (i.e. ln(s)) Laplace’s equation in this case is, ∇2V = 1 s ∂ ∂s (s ∂V ∂s ) = 0; so s ∂V ∂s = b (12) where b is a constant. Integrating we find, V (s) = a+ bln(s), which is the general solution in the cylindrical case. In spherical polar co-ordinates, ∇2V = 1 r2 ∂ ∂r (r2 ∂V ∂r ) = 0; so r2 ∂V ∂r = b (13) Integrating, we find the general solution V (r) = a + b/r, which is the potential outside charge distributions with spherical symmetry. The one variable solutions to Laplace’s equation are monotonic i.e. decreasing or increasing with no minima or maxima on their interior. Nevertheless electrostatic potential can be non-monotonic if charges are added as a change in slope can occur at charges. However if a charge is included, the one dimensional solutions to Laplace’s equation only apply to the regions of space that do not have charge. They do NOT apply in regions where there is charge, for example inside a sphere with a constant charge density. In that case we have to solve Poisson’s equation, for example in the case of a sphere with constant charge density we solve, ∇2V = 1 r2 ∂ ∂r (r2 ∂V ∂r ) = − ρ 0 ; (14)