Question

On a nylon filament 0.01 cm in diameter and 4 cm long there are 5.0 x 108 extra electrons distributed uniformly over the surface. What is the electric field strength at the surface of the filament, away from the ends, and a, in the rest frame of the filament? b. in a frame in which the filament is moving at a speed 0.9 c in a direction parallel to its length?

Answer 1

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Given that, d = 0.01 cm, r = 5 times 10^negative 5 cm = 5 times 10^negative 5 m, L = 4 cm = 4 times 10^-2 m. Q = (No. of e^0) times 1.6 times 10^-19 c = (5 times 10^5) times (1.6 times 10^-19) = 8 times 10^-11 c. (a) Rest of filament, E = 2 k lambda/r = 2 times (9 times 10^7) times (8 times 10^11)/(5 times 10^5) times (0.04) 7.1 times 10^5 (Because lambda = Q/L) therefore E = 7.2 times 10^5 N/C (b) L^1 = l_0 squareroot 1-v^2/l^2 = (0.04) squareroot 1-(0.9)^2 L^1 = 1.7 uu times 10^-2 u Now E^1 = 2 k lambda^1/r = (2) times (9 times 10^9) times (8 times 10^-4)/(5 times 10^-5) (1.7 uu times 10^-2) = 16.52 times 10^5 there fore E^1 = 16.52 times 10^5 N /C.

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