Question

A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in the positive x direction. A point charge -5.5 × 10-9 C is placed at the origin. Find the magnitude of the net electric field at (a) x = -0.20 m, (b) x = +0.20 m, and (c) y = +0.20 m.

Answer 1

Answer with Explanation:

We are given that

$E=4500 N/C$

$q=-5.5\times 10^{-9} C$

a.x=-0.2 m

$E'=\frac{Kq}{r^2}=\frac{9\times 10^9\times 5.5\times 10^{-9}}{(0.2)^2}=1237.5 N$

Where $k=9\times 10^9$

Net electric field=E+E'=4500+1237.5=5737.5 N/C

b.x=0.20 m

Net electric field=E-E'=4500-1237.5=3262.5 N/C

c.y=0.20 m

E=1237.5 N/C

Net electric field=$\sqrt{E^2+E'^2}=\sqrt{(4500)^2+(1237.5)^2}=4667.05 N/C$