Answer to Question #91784 in Electricity and Magnetism for Alexa

Question

Point P is located at the origin of a coordinate plane.
Object A of charge 2e-9 C is located on the x-axis of a coordinate plane at x = 0.4 m.
Object B of charge -4e-9 C is located on the y-axis of a coordinate plane at y = 0.5 m.
Calculate the direction (angle formed with the positive x-axis) of the net
electric field at point P due to objects A and B.

-46.801°
153.598°
-52.001°
127.999°
115.199°

Point P is located at the origin of a coordinate plane. Object A of charge -5e-9 C is located on the x-axis of a coordinate plane at x = 0.1 m. Object B of charge -2e-9 C is located on the y-axis of a coordinate plane at y = 0.2 m.Calculate the magnitude of the net electric field at point P due to objects A and B.

4522.444 N ⁄ C
5879.177 N ⁄ C
4070.2 N ⁄ C
2713.466 N ⁄ C
5426.933 N ⁄ C

Answer 1

1) Calculate the angle. First, determine the field at point A on the x-axis:

$E_A=\frac{kq_A}{r_A^2}=112.34\text{ V/m}.$

At point B on the y-axis:

$E_B=\frac{kq_B}{r_B^2}=-143.80\text{ V/m}.$

Hence, the angle will be

$\alpha=90^\circ+\text{atan}\frac{112.34}{143.80}=128^\circ.$

2) The magnitude of the vector can be calculated the same way:

$E_A=\frac{kq_A}{r_A^2}=-4493.78\text{ V/m}.$

$E_B=\frac{kq_B}{r_B^2}=-449.38\text{ V/m}.$

these vectors are at the right angle to each other, so the total vector at point P is

$E=\sqrt{E_A^2+E_B^2}=4516.19\text{ V/m}.$