Question

2. A thin ring of radius R in the x - y plane is centered at the coordinate origin, and is charged with linear charge density λ which depends* on the polar angle θ as (9) λο sin(0), where 0 > 0, and θ . (a) Plot λ(0) for θ [0.2n]. (b) Before doing any calculations, sketch the electric field vector vector at the coor- - 0 is on the positive r-axis dinate origin in the direction you expect it to point. [Hint 1: keep in mind the orientation of the electric field - it points away from positive charges (c) Calculate the components E and Ey of the electric field at the coordinate origin.

Answer 1

Signs of charge as sin theta +ve in 1st & 2nd quadrant and -ve in rest quadrants. (a) lambda(theta) = lambda_0 sin theta theta elementof[0, 2 pi] (b) sin(-theta) = -sin theta therefore theta clockwise or anticlockwise will have equal magnitude but opposite sign and so will change have. for a point P making angle theta and P' making angle -theta with x-axis, electric field is shown in figure. above on top. X-component cancels and adds in y component. This would be true for all pairs thus E_net in -ve y axis direction. \r\n (c) As discussed in part (b) E_x = 0 E_y = ? consider fig 1 and point P and P' Electric field at origin due to dq is dE = k dq/R^2 = k(lambda_0 sin theta)R d theta/R^2 Now as x-components cancel and y adds dE_net = 2 dE sin theta = 2 times k(lambda_0 sin theta)R d theta times sin theta/R^2 = 2 k lambda_0/R