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?

Respuesta :

Answer:

57.94°

Explanation:

we know that the expression of flux

[tex]\Phi =E\times S\times COS\Theta[/tex]

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 [tex]\frac{Nm^{2}}{sec}[/tex]

                          E=[tex]1.44\times 10^{4}\frac{Nm}{C}[/tex]

                          S=[tex]\pi \times 0.057^{2}[/tex]

                         [tex]COS\Theta =\frac{\Phi }{S\times E}[/tex]

 =   [tex]\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}[/tex]

 =0.5306

 θ=57.94°