Question

Q1. A curved plastic rod of charge+Q forms a semi-circle of radius R in the x-y plane, as shown below on the left. The charge is distributed uniformly across the rod. dQ +Q +Q Now let's analytically determine the magnitude and direction of the electric field E at the center of the circle using polar coordinates and the charge element dQ shown in the image on the right Write down an expression for the electric field dE at the center of the circle due to this element dQ. Hints: a. a. Remember, dE is a vector, write using the l,J notation. b. your answer should be written in terms of R, use that dQ = λ dl where now λ is the linear charge density λ = 요 (total charge/divided by total length) use polar coordinates where dl Rde c. d. b. Write down the integral you need to do to solve for E,and E, at the center of the semi-circle (be

Answer 1

dt = length of elementary part lambda = linear charge density = Q/L here = dl = R d theta dQ = lambda dl Therefore From the diagram we can see that dE = dQ/4 pi elementof_0 R^2 Therefore dE vector = -dQ/4 pi elementof_0 R^2 cos theta i^^- dQ/4 pi elementof_0 R^2 sin theta j^^or, dE vector = -dQ/4 pi elementof_0 R^2 [cos theta i^^+ sin theta j^^] or, dE vector = -lambda dl/4 pi elementof_0 R^2 [cos theta i^^+ sin theta j^^] From the diagram it is clear that dE cos theta components will cancel out for the whole semicircular ring. Therefore E = integral dE = integral - Q/4 pi elementof_0 L R^2 sin theta dl = integral - Q/4 pi elementof_0 L R^2 times sin theta times R d theta = -Q/4 pi elementof_0 L R^2 integral^pi_0 sin theta d theta \r\n E = -Q