Question

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 31.6° above the horizontal?

Answer 1

Answer:231.16 N/C

Explanation:

Given

Electric Flux$=147 N-m^2/C$

Area(A)$=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6=$58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C