Question

A flat sheet with an area of 3.8 m 2 is placed in a uniform electric field of magnitude 10 N/C. The electric flux through the sheet is 6.0 Nm 2 /C . What is the angle (in degrees) between the electric field and sheet's normal vector?

Answer 1

The angle between the electric field and sheet's normal vector is 80.96 degrees.

Step-by-step explanation:

Given that,

Area of the flat sheet,
`A=3.8\ m^2`

Electric field, E = 10 N/C

Electric flux of the sheet,
`\phi=6\ Nm^2/C`

The electric flux is through the sheet is given by the dot product of electric field and the area vector. It is given by :

`\phi=E{\cdot} A`

or

`\phi=EA\ cos\theta`

`\theta` is the angle between electric field and sheet's normal vector

So,

`cos\theta=(\phi)/(EA)`

`cos\theta=(6)/(10* 3.8)`

`\theta=cos^(-1)(0.157)`

`\theta=80.96^(\circ)`

So, the angle between the electric field and sheet's normal vector is 80.96 degrees. Hence, this is the required solution.