Question

Problem 3: the infinite cylinder An insulating cylinder that is infinitely long has radius R and a charge per unit length of λ. (Hint: because it is an insulator you should assume that the charge is spread uniformly across its entire volume of the cylinder) a) Use Gauss' Law to calculate the electric field at a point outside of the cylinder as a function of r, the radial distance from the center of the cylinder. (r> R) b) Use Gauss' Law to calculate the electric field at a point inside the cylinder as a function of r, the radial distance from the center of the cylinder. (r< R) c) Make a plot of the electric field magnitude E(r) as a function of r for any r >0.

Answer 1

The infinite cylinder a) Outside the cylinder (r > R) The flux through the Gaussian surface is phi = contourintegral E.da = E (2 pi r l) The charge enclosed by the gaussian surface is q_0n = integral^r_0 P dv < Because p = lambda > = integral^R_0 lambda dv + integral^r_R (0) dv = integral^R_0 lambda dv < Because lambda is uniform > = lambda integral^R_0 dv = lambda integral^R_0 r dv d phi dz = lambda integral^R_0 r dv integral^2pi_0 d phi integral^l_0 dz = lambda (r^2/2)^R_0 (2 pi) (l) = lambda R^2/2 times 2 pi l q_en = lambda pi l R^2 Therefore From Gauss's law, we have phi = q_ec/epsilon_0 E(2 pi r l) = lambda pi l R^2/epsilon_0 E = lambda R^2/2 epsilon_0 r implies E alpha 1/r b) Inside the cylinder (r < R) The flux through the gaussian surface is phi = contourintegral E.da phi = E(2 pi r l) The charge enclosed by the gaussian surface is..