Electrostatic Potential of a Line Charge Near a Grounded Conducting Cylinder - Prof. Phill, Study notes of Physics

Download Electrostatic Potential of a Line Charge Near a Grounded Conducting Cylinder - Prof. Phill and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 13 A line charge near a grounded conducting cylinder A line charge λ at position x0 on the x-axis is near a grounded conducting cylinder, of radius R, which has its center at (x, y) = (0, 0), with its central axis along the z-axis. The line charge lies parallel to the central axis of the cylinder. Find the electrostatic potential for r > R in plane cylindrical co-ordinates r, φ. Note that the potential does not depend on z. Using Gauss’s law it is easy to show that the electric field near a uniform line charge is ~E(r) = λr̂/(2πǫ0r). The potential is then of the form V (r) = λln(constant/r)/(2πǫ0). The constant is chosen to fit the boundary conditions. Let’s assume that the problem of a line charge near a grounded conducting cylinder is solved by using an image charge which is located at position x′ 0 and with charge per unit length −λ. Now we need to find x′ 0 and we need to check that V (r, φ) = 0, and that ~E(R, φ) = Ern̂. The potential is given by superposition, so that, V (r, φ) = λ 2πǫ0 [ln(c1/r1) − ln(c1/r2)] + c2 (1) where c1 and c2 are constants. Using the cosine rule we have, r2 1 = r2 + x2 0 − 2rx0cosφ; r 2 2 = r2 + x′ 0 2 − 2rx′ 0 cosφ (2) Lets also assume that the reciprocal relation holds (why not!!), ie x′ 0 = R2/x0. We then have V (r, φ) = λ 4πǫ0 ln( r2 + R4/x2 0 − 2r(R2/x0)cosφ r2 + x20 − 2rx0cosφ ) + c2 (3) At the surface of the cylinder we have, V (R, φ) = λ 4πǫ0 ln([ R2 x20 ]( R2 + x2 0 − 2Rx0cosφ R2 + x20 − 2Rx0cosφ )) + c2 (4) This must be zero for our solution to be correct, which implies that, c2 = − λ 2πǫ0 ln(R/x0) (5) The solution to our problem is then, V (r, φ) = λ 4πǫ0 ln( r2 + R4/x2 0 − 2rR2/x0cosφ r2 + x20 − 2rx0cosφ ) − λ 2πǫ0 ln(R/x0) (6) 1 or V (r, φ) = λ 4πǫ0 ln( x2 0 r2 + R4 − 2rx0R 2cosφ r2 + x20 − 2rx0cosφ ) − λ 2πǫ0 ln(R) (7) Now we need to check that the electric field is given correctly. We find that, Eφ = −1 r ∂V ∂φ = −λ 4πǫ0r ( 2rx0R 2sinφ x20r 2 + R4 − 2rx0R2cosφ − 2rx0sinφ r2 + x20 − 2rx0cosφ ) (8) From Eq. (7) it is evident that V (R, φ) = 0 as required and from Eq. (8) we find Eφ(R, φ) = 0. We have therefore found a solution which satisfies the boundary condition, so it is correct. For completeness, the electric field in the radial direction is given by, Er = − ∂V ∂r = −λ 4πǫ0 ( 2rx2 0 − 2x0R 2cosφ x20r 2 + R4 − 2rx0R2cosφ − 2r − 2x0cosφ r2 + x20 − 2rx0cosφ ) (9) Closing remarks on generalizing image charge problems We have solved three basic image charge problems: (i) a point charge near a grounded flat conducting surface (Lecture 12) (ii) a point charge near a grounded conducting sphere (Lecture 12) (iii) a line charge near a grounded conducting cylinder (This Lecture). Lets call these solutions VG(~r). The extension to problems where the conductor is at some finite voltage (instead of zero) requires adding charges to produce that voltage. The charges have to be placed symmetrically to ensure that no electric field is generated in the metal. E.g. if we want a sphere of radius R at potential V0, then we place an image charge Q0 at the center of the sphere so that V0 = kQ0/R. This corresponds to distributing the charge Q0 uniformly on the surface of the sphere. The electrostatic potential for r > R of this problem is found by superposition, i.e. V (~r) = VG(~r)+ kQ0/r. In the case of a conducting slab, a sheet of image charge is placed at the center of the slab, while in the case of a conducting cylinder, a line charge is placed at the center of the cylinder. In a similar way, if we are given a problem where a point charge is near an isolated conducting sphere, cylinder or slab which has total charge Q, then we again have to place an image charge at the center of the sphere. However, now the magnitude of the image charge at the center of the metal sphere has to be the sum of the total charge on the sphere plus the value of the image charge of the grounded system. For example an isolated conducting sphere of charge Q requires that an image charge of Q − q′ be placed at its center, so the total potential for r > R becomes VG(~r) + k(Q − q ′)/r, where q′ = −qR/z0 is the image charge of the grounded sphere. 2