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

Answer:

$\phi = 1.28 Nm^2/C$

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

$\theta = 90 - 35 = 55$

$\phi = EAcos\theta$

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

$\phi = 1.28 Nm^2/C$