Answer to Question #298471 in Electric Circuits for Jc Bondad

Question

A +40uC charged is placed at the origin. Calculate the magnitude and direction of the field created by the point charged at the following A point (5m,0) B. (3m,4m)

Answer 1

Part A

$E=\frac{kq}{r^2}\\r=5m$

$\therefore\>E=\frac{9×10^9×40×10^{-6}}{5^2}$

$=1.44×10^4N/C$

Toward positive x-axis

Part B

$r=\sqrt{3^2+4^2}=5m$

$|E|=1.44×10^4N/C$

$\theta=$ tan$^{-1}\>\frac{4}{3}=53.13$

$E=1.44×10^4N/C$

Along a line 53.13° to x-axis directed away from origin.