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}*20\text{ cm}`

Required: calculate the flux passing through the surface

Step-by-step 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* A`

Plugging all the values in the above relation, we get

`\begin{gathered} \phi=400\text{ N/C}*\cos60^(\degree)*30\text{ cm}*20\text{ cm} \\ \phi=400\text{ N/C}*(1)/(2)*0.3\text{ m}*0.2\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}`