Question

(a) A thin plastic rod of length L carries a uniform linear charge density, λ-20 trCm, along the x-axis, with its left edge at the coordinates (-3,0) and its right edge at (5, 0) m. All distances are measured in meters. Use integral methods to find the x-and y-components of the electric field vector due to the uniformly-charged charged rod at the point, P. with coordinates (0, -4) m. 4, (o, 4 p2212sp2018 tl.doex

Answer 1

As shown in figure if p is my general point we considered an element on rod of length dx at a distance x from 0 as shown in figure 2 now de be the electric field at p due to element then de = kdq/(x^2 + r^2) (dq = (theta/L) dx = lambda dx) electric field strength in x direction due to dq De = De sin theta = [kdq/(x^2 + r^2) sin theta = l lambda/(x^2 + r^2) dx here x = r tan theta and dx = r sec^2 theta d theta d E_x = k theta/L integral_-theta_2^theta_1 sin theta d theta = k l/r [-cos theta]_-theta_2^theta_1 E_x = k l/r [cos theta_2 - cos theta_1] equation (1) similarly Electric field strength at point p due to dq y direction as d E_y = DE cos theta De_Y = k lambda dx/(r^2 + x^2) cos theta we have x = r tan theta implies dx = r sec^2 theta