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

57.94°

Step-by-step explanation:

we know that the expression of flux

`\Phi =E* S* 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
`(Nm^(2))/(sec)`

E=
`1.44* 10^(4)(Nm)/(C)`

S=
`\pi * 0.057^(2)`

`COS\Theta =(\Phi )/(S* E)`

=
`(78)/(1.44* 10^(4)* \pi * 0.057^(2))`

=0.5306

θ=57.94°