Question

1.16 Gauss's law and two point charges* (a) Two point charges q are located at positions x = ±1. At points close to the origin on the x axis, find Ex. At points close to the origin on the y axis, find Ey. Make suitable approximations with x <E and y< e (b) Consider a small cylinder centered at the origin, with its axis along the x axis. The radius is ro and the length is 2x0. Using your results from part (a), verify that there is zero flux through the cylinder, as required by Gauss's law.

electric field and flux

Answer 1

(a) point p is at (x, 0) x < < l E_x = 1/4 pi elementof_0[q/(f + x)^2 - q/(l - x)^2] E_x = 1/4 pi elementof_0[q[l - mu)^2 - (l + x)^2]/(l^2 - x^2)^2] E_x = 1/4 pi elementof_0[- 49 lx/(l^2 - x^2)^2] E_x = 1/4 pi elementof_0 [- 4q lx/l^4(1 - x^2/l^2)^2]almostequalto - 49 x/4 pi elementof_0 l^3 E_1 = E_x i + E_y j E = E_1 + E_2 = 2 E_y j E_y = q/4 pi elementof_0[2y/(l^2 + y^2)^3/2] = 2qy/4 pi elementof_0 l^3(1 + y^2/l^2)^3/2 E_y almostequalto 2qy/4 pi elementof_0 l^3 (b) Radius of cylinder = r_0 length of cylinder = 2x_0 x = x_0 y = r_0 since cylinder is small, E_y is uniform over surface of cylinder and E_X is uniform over flat ends of cylinder contourintegral_s E ds = integral_ends E ds + integral_sides E ds = (pi r^2_0) (- 4q x_0/4 pi elementof_0 l^3)(2) + 2 pi r_0 (2x_0) (2qr_0/4 pi elementof_0 l^3) = -8 pi r^2_0