The electric intensity on the surface of a charged conductor of area 0.5 `m^(2)` is 200 V/m. if the electric flux is 86.6 `Nm^(2)//C`, then the angle between the normal drawn to the surface and the electric intensity is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

B

Explanation

`phi=ES cos theta" "costheta=(phi)/(E_(S))=(86.6)/(200xx0.5)=0.866`
`therefore theta=30^(@)" "[cos30^(@)=(sqrt(3))/(2)=0.866]`