A-Level Coulomb's law, electric field strength, potential and energy, uniform fields and capacitance: worked solution

Question

(ii) Using your result from (a)(i), apply Gauss's law with a suitably chosen Gaussian surface to show that, for z ≪ R, the electric field at the point (0,0,z) is approximately Ez = 2πkσ where k is a constant you must find. A sketch may help your working.

For z << R the charged surface (density sigma) looks effectively infinite, so the field near (0,0,z) is uniform and perpendicular to the surface. Choose a Gaussian 'pillbox': a short cylinder of face area A straddling the surface, faces parallel to it. By symmetry flux only exits the two faces: Phi = 2 E A. Charge enclosed = sigma A. Gauss's law: 2 E A = sigma A / epsilon0, so E = sigma / (2 epsilon0). Writing the Coulomb constant k = 1/(4 pi epsilon0), i.e. epsilon0 = 1/(4 pi k): E = sigma / (2 * 1/(4 pi k)) = 2 pi k sigma. Hence Ez = 2 pi k sigma with k = 1/(4 pi epsilon0).