Question

An electric field has a magnitude of 400 N/C and makes an angle 60o with the perpendicular to the surface of a flat surface of dimensions 30 cm by 20 cm. Calculate the electric flux through the surface.please kindly help me out.

Answer 1

Given:

the value of electric field is

$E=400\text{ N/C}$

The field makes an angle with surface is

$\theta=60^{\degree}$

The area of the surface is

$A=30\text{ cm}\times20\text{ cm}$

Required: calculate the flux passing through the surface

Explanation: look at the free body diagram

flux passing through the surface is given by

$\phi=EA$

here,

$E$

is the electric field and

$A$

is the area of the surface

from the figure we can write it as

$\phi=E\cos\theta\times A$

Plugging all the values in the above relation, we get

$\begin{gathered} \phi=400\text{ N/C}\times\cos60^{\degree}\times30\text{ cm}\times20\text{ cm} \\ \phi=400\text{ N/C}\times\frac{1}{2}\times0.3\text{ m}\times0.2\text{ m} \\ \phi=12\text{ N m}^2\text{ /C} \end{gathered}$

Thus, the flux passing through the surface is

$12\text{ N m}^2\text{ /C}$