Question

An electric field is constant at every point on a square surface that is 0.80 m on a side. This field has a magnitude of 3.5 N/C and is oriented at an angle of 35° with respect to the surface, as the drawing shows. Calculate the electric flux ΦE passing through the surface.

Answer 1

`\phi = 1.28 Nm^2/C`

Step-by-step explanation:

As we know that electric flux is defined as

`\phi = \vec E . \vec A`

now we have

`A = L^2`

`A = 0.80^2 = 0.64 m^2`

`E = 3.5 N/C`

also we know that electric field makes and angle of 35 degree with surface

so angle made by electric field with Area vector is given as

`\theta = 90 - 35 = 55`

`\phi = EAcos\theta`

`\phi = (3.5)(0.64)cos55`

`\phi = 1.28 Nm^2/C`