Question

Problem 6. [20] A very long cylindrical capacitor has an inner conductor of fadius and an outer conductor of radius b. The two conductors are separated by vacuum. (a) First consider only the inner cylinder with radius a and charge per unit length (ie, ignore the outer conductor for this part). From Gauss's law, one can show that the electric field outside the cylinder at distance r from its axis has only a radial component and is given by: E-wl2m) Using this, derive an expression for the potential, Vr), outside the inner conductor, setting V-0 atrb (with b a). Express your answer in terms of a, b, r, i and relevant constants. (S points) (b) Now consider both the inner conductor with charge per unit length à and the outer conductor with charge per unit length -i and again set V-0 atr-b. Find and draw Vr), including regions 0-rca, acrb, and r>b. (5 points) (c) Using your answer above, calculate the capacitance per unit length. Give your answer in terms of a, b, and relevant constants. (5 points) (d) If the potential of the inner conductor relative to that of the outer conductor is Vab -500 mV, a 3.0 mm, b 4.0 mm, and the capacitor's length is L-4.0 m, find the charge (magnitude and sign) on the inner conductor. (5 points)

Answer 1

Given E(r) = lambda/2 pi r epsilon_0 V(r) = -integral^r_infinity E dr V(r) = -lambda/2 pi epsilon_0 integral^r_infinity 1/r dr V(r) = -lambda/2 pi epsilon_0 [ln[r]] V(r) = -lambda ln(r)/2 pi epsilon_0 \r\n potential when r > b lambda_T = 0 electric field = 0 V_2 = potential = 0 a < r < b V_1 = -integral^r_b lambda/ 2 pi epsilon_0 r dr = lambda/2 pi epsilon_0 [-ln[r]]^r_b V(r) = lambda/2 pi epsilon_0 ln(b/r) r < a V_0 = -integral^a_b lambda/2 pi epsilon_0 r dr V_0 = lambda/2 pi epsilon_0 ln(b/a) \r\n capacitance C = Q/V = Q/lambda/2 pi epsilon_0 ln(b/a) = lambda l 2 pi epsilon_0/lambda ln(b/a) = 2 pi l epsilon_0/ln(b/a) V_ab = lambda/2 pi epsilon_0 ln(b/a) and Given L = 4 m a = 3 mm b = 4 mm V_ab = -500 m vott Q = ?? \r\n so lambda = V_ab 2 pi epsilon_0/ln(b/a)