Finding the Electric Field - Electricity and Magnetism I | PHY 481, Study notes of Physics

Download Finding the Electric Field - Electricity and Magnetism I | PHY 481 and more Study notes Physics in PDF only on Docsity! PHY481 - Lectures 7 and part of 8 Sections 3.5-3.6 of PS A. Finding the electric field - continued (vi) Uniform shell of charge (shell theorems) The shell theorems state that (i) the electric field inside a uniform shell of charge is zero and (ii) that the field outside the uniform shell of charge is the same as that of a point charge with the same total charge as the shell. These results are easy to derive from Gauss’s law, in the same way as for the cylindrical case. However now we use spherical surfaces of radius r < R and r > R, where R is now the radius of the spherical shell of charge. The shell has charge density σ and the total charge on the shell is Q = 4πR2σ. We first consider a Gaussian surface consisting of a spherical shell, which is concentric with the charged shell, with radius r < R. Noting that by symmetry the electric field is directed radially and only depends on r, then since there is no enclosed charge the electric field for r < R is zero in agreement with the first shell theorem. Considering a spherical shell of radius r > R, we have, ∫ S ~E · ~dA = 4πR2E(r) = Q ǫ0 so that ~E = 1 4πǫ0 Qr̂ r2 (1) which proves the second shell theorem. Now lets prove the first shell theorem directly. Consider a general point ~r inside the sphere. If we place a spherical co-ordinate system at that point, then we can draw a cone of infinitesimal solid angle dΩ = sinθdθdφ extending from this origin to the spherical surface. The cone has two intersections with the spherical shell, on opposite sides of the sphere. The areas of the spherical shell subtended by this solid angle are dA1 = r 2 1 dΩ and dA2 = r 2 2 dΩ. The magnitude of the electric field at ~r due to these two areas of charge is given by, dE = kσ( dA1 r21 − dA2 r22 ) = 0 (2) This construction applies to all parts of the spherical surface, proving the first shell theorem. Proof of the second shell theorem is more involved. It is also relatively straightforward to demonstrate the second shell theorem by direct integration, using polar co-ordinates it may be shown that the electric field in the radial direction for r > R where R is the radius of the shell, is given by, Er(r) = σ 4πǫ0 ∫ π 0 (r − Rcosθ)2πR2sinθdθ (R2sin2θ + (r − Rcosθ)2)3/2 (3) 1 which reduces to the point charge result. in deriving this expresion we used dA = 2πR2sinθdθ (vii) Uniformly charged sphere of radius R Again we carry out the flux integral to find, for a surface of radius r, φE = E(r)4πr 2 (4) The charge enclosed by this surface changes with r. For r < R, the enclosed charge is given by, q(r < R) = 4π 3 r3ρ (5) while for r > R we have, q(r > R) = 4π 3 R3ρ (6) Using Gauss’s law φE = q/ǫ0, we thus have, E(r < R) = rρ 3ǫ0 = Q 4πǫ0 r R3 (7) where Q = 4πρR3/3 is the total charge on the sphere. The electric field outside the sphere of charge is like that of a point charge (shell theorem), E(r > R) = 1 4πǫ0 Q r2 (8) In both cases, the direction of the electric field is r̂. (see problem 3.19) B. Electric potential energy and electric potential Physical definition The electric potential energy (U) is the potential energy due to the electrostatic force. As always only differences in potential energy correspond to physical observables. However we define a reference potential energy and calculate all differences in potential energy with respect to this reference. In electrostatics, the potential energy is defined to be zero when the charges are an infinite distance apart. The difference in potential energy in moving a charge between two positions a and b is defined in terms of the work done in moving the charge between these two positions, so that, ∆Uab = ∫ b a ~Fext · ~ds = − ∫ b a ~F · ~ds (9) 2