Question

Problem A.1 - Calculate electric flux f5) The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E . Consider an imaginary cylinder with radius e-25 cm and length L = 40 cm that has an infinite line of positive charge running along its axis. The charge per unit length is 3 HC/m. Do not use Gauss's Law, but actually calculate the flux! a) What is the electric flux through the cylinder due to this infinite line of charge? 3) b) What is the flux through the cylinder if its radius is increased to e = 50 cm? {1} c) What is the flux through the cylinder if its length is increased to L-80 cm? Problem A.2 - Gauss's Law (5) An infinite cylinder of charge has radius a and uniform charge per unit length λ. a) Make a sketch of the problem and your choice of Gaussian surface. Indicate where the enclosed charges are located. Use Gauss's Law to find the electric field inside and outside the cylinder of charge. 3) inder. Again, use Gauss's Law to ascertain the electric field in every region. ) unit volume ps, centred on the original cylinder. Use Gauss's Law to ascertain the electric field in every region. 1) b) We add a thin infinite cylindrical shell of radius b > a, and uniform charge per unit area σ-, Centred on the previous cy- c) We add to the cylinder and shell an infinite cylindrical shell of radius c > b of finite width Δε and uniform charge per

Answer 1

E = lambda/2 pi E_0 R a) phi = E A cos theta A = 2 pi r l (cylinder) phi = lambda/2 pi E_0 r (2 pi r l) phi = lambda l/E_0 = 3 times 10^-6 times 40 times 10^-2/8.2.25 times 10^-12 = 13.59 times 10^4 NM^2/L (2) Radius of cylinder r Electric flux phi = E A cos theta theta = 0 phi = lmabda/2 pi E_0 r (2 pi r l) = lambda l/E_0 = 3 times 10^-6 times 40 times 10^-2/2.25 times 10^-12 phi = 13.53 times 10^4 Nm^2/C (3) Length of cylinder l = 0.80 m phi = lambda l/E_0 = 3 times 10^-6 times 30 times 10^-2/2.25 times 10^-12 = 27.11 times 10^4 Nm^2/C