More on Problems Solved Using Image Charge Methods | PHY 481, Study notes of Physics

Download More on Problems Solved Using Image Charge Methods | PHY 481 and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 12 Sections 4.2, 4.3 of PS A. More on problems solved using image charge methods First lets solve the simplest image charge problem completely using polar co-ordinates. The geometry is a grounded semiinfinite metal with normal along positive k̂ and metal for all z < 0. A charge q is placed on the z-axis at position z0. Find the potential for z > 0. From this expression show that the potential at z = 0 is zero. Find the electric field and show that at z = 0 the electric field is directed along the z-axis. Using the image charge method, we have, V (r, θ) = kq[ 1 (r2 + z20 − 2rz0cosθ) 1/2 − 1 (r2 + z20 + 2rz0cosθ) 1/2 ] (1) The surface z = 0 corresponds to θ = π/2. At this value of θ it is easy to show that the potential. The electric field in polar co-ordinates is given by, ~E = − ∂V ∂r r̂ − 1 r ∂V ∂θ θ̂ − 1 rsinθ ∂V ∂φ φ̂ (2) The φ̂ component is zero as there is no φ dependence in the potential. Doing the derivative for the other two components yields, Eθ = kq[ z0sinθ (r2 + z20 − 2rz0cosθ) 3/2 + z0sinθ (r2 + z20 + 2rz0cosθ) 3/2 ] (3) and Er = kq[ r − z0cosθ (r2 + z20 − 2rz0cosθ) 3/2 − (r + z0cosθ) (r2 + z20 + 2rz0cosθ) 3/2 ] (4) Now we need to show that these equations satisfy the required boundary conditions that V (z = 0) = 0 and that ~E(z = 0) = −Eθ(z = 0)n̂, that is the potential at the surface is zero and the electric field is normal to the surface. In polar co-ordinates, the surface is at V (r, π/2). Evaluating this demonstrates that V (r, π/2) = 0 as required. To show that the electric field is normal to the surface we need to show that Er(r, π/2) = 0 and Eθ(r, π/2) is finite. These results follow from Eq. (3) and (4), moreover, we find that, Eθ(r, π/2) = 2kqz0 (r2 + z20) 3/2 (5) Finally we would like to show that the total induced charge at the surface of the metal is −q to do that we use the fact that at a metal surface, ~E = σn̂/ǫ0, which follows from Gauss’s 1 law. The induced charge at the metal surface is given by, Q = ∫ σdA = − ∫ ∞ 0 2πrdrǫ0Eθ = − ∫ ∞ 0 2πrdrǫ0 2kqz0 (r2 + z20) 3/2 = −q (6) The minus sign in the second expression arises as the θ direction is −n̂ for this surface. Some problems can be solved using many image charges, one example where four image charges works is the case of a charge q at (a, a, 0) when the regions y < 0 and x < 0 are conducting and grounded. In that case three image charges −q at (a,−a, 0), q at (−a,−a, 0) and −q at (−a, a, 0) ensure that the potential on the surfaces of the metal regions are all zero. In a similar way, the method of images can be used for a wedge of angle π/n (the case above has wedge angle π/2) when a charge is placed on the central axis of the wedge. In that case alternating image charges placed at symmetric positions at the centers of all wedges inside the metal sum perfectly to ensure that the potential is zero on the wedge surfaces. In some cases an infinite set of image charges can be used with one example being a charge lying between two grounded metal sheets at locations z = d and z = −d. The solution is found by adding image charges iteratively to find, V (r) = kq[ 1 r + ∞∑ k=1 2(−1)k (r2 + 4d2k2)1/2 ] (7) This can be generalized to the non-symmetric case by using a similar procedure. This approach can also be used for some problems involving two spheres or two cylinders. Another case which can be solved using one image charge is the case of a charge outside a grounded spherical conductor. Now we can try to solve the problem in the same way as for the flat surface but now using spherical co-ordinates centered at the sphere center. There is azimuthal symmetry so we can restrict consideration to a function V (r, θ, and we have the boundary condition V (R, θ) = 0 as the sphere is grounded. The point charge is at z0 > R. In this case we need to allow the image charge to have a general location and a general charge so that, V (r, θ) = k[ q (r2 + z20 − 2rz0cosθ) 1/2 + q′ (r2 + z′20 − 2rz ′ 0cosθ) 1/2 ] (8) Clearly there are two unknowns which can be determined by choosing two convenient loca- tions, for example θ = 0, π, r = R. These two cases imply that, kq R − z0 ± kq′ R − z0 = 0; kq R + z0 ± kq′ R + z0 = 0; (9) 2