Question

A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The magnitude of the electric flux through the surface is 78 N · m2/C. What is the angle (less than 90) between the direction of the electric field and thenormal to the surface?

Answer 1

Answer:

57.94°

Explanation:

we know that the expression of flux

$\Phi =E\times S\times COS\Theta$

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 $\frac{Nm^{2}}{sec}$

                          E=$1.44\times 10^{4}\frac{Nm}{C}$

                          S=$\pi \times 0.057^{2}$

                         $COS\Theta =\frac{\Phi }{S\times E}$

 =   $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

 =0.5306

 θ=57.94°