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

Answer 2

The electric flux depends on the electric field and area of the surface also the angle between the area vector and electric field.

The magnitude of the electric field is 341.86 N/C.

What is an electric field?

An electric field can be defined as a field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

Given that 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 field points 31.6° above the horizontal.

The angle between the area vector and electric field is,

$\theta = 90^\circ - 31.6^\circ$

$\theta = 58.4^\circ$

The flux is given as,

$\phi = EAcos\theta$

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

$E = 341.86\;\rm N/C$

Hence we can conclude that the magnitude of the electric field is 341.86 N/C.

To know more about the electric field, follow the link given below.

https://brainly.com/question/12757739.