Reference. Physics of Scientist and Engineers 6th edition 3.- (10 points) An inf

Question

Reference. Physics of Scientist and Engineers 6th edition

3.- (10 points) An infinite line of charge (linear charge density lambda = +2 nC/m) lies parallel to the y axis at x = -1m. An infinite plane of charge perpendicular to the xy plane (surface charge density sigma = +4 nC/m^2) cuts the line charge at (-1m, -1m) forming an angle a = 450 with the line. Find the electric field vector E at P (0,1m)? Hint: Let 0 degree correspond the direction parallel to the x- axis. Use the formula 22-6 for the electric field due to a uniformly charged line and 22-10 for the electric field of a uniform plane charge.

Explanation / Answer

electric field at P due infinite line of charge

E1x = lambda/*2*pi*eo*x)

x = perpendicular distance from the line of charge = 1 cm = 0.01 m

E1x = (2*10^-9)/(2*3.14*8.85*10^-12*0.01)

E1x = 3600 N/C

electric field due to infinite plane of charge

E2 = sigma/2eo   perpendicular to plane

E2x = -E2*cos45 = (4*10^-9*cos45)/(2*8.84*10^-12) = -160 N/C

E2y = +E2*sin45 = (4*10^-9*sin45)/(2*8.84*10^-12) = 160 N/C

Ex = E1x + E2x = 3440 N/C

Ey = 160 N/C

E = (3440 , 160 ) <-answer

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