Question

​ A flat surface with area of 7.00 m 2 is oriented at an angle θ to a uniform electric field with magnitude 6.0 × 10 5 N/C. If the flux through this surface is 2.0 × 10 6 N ⋅ m 2 /C, what is θ ?

Answer 1

Answer:
`\theta =28.42^(\circ)`

Step-by-step explanation:

Given

Area of surface
`A=7\ m^2`

Electric Field strength
`E=6* 10^5\ N/C`

Flux through Surface
`\phi =2* 10^6\ N-m^2/C`

Surface is Oriented with Electric field at an angle
`\theta`

so Angle between Area vector and Electric Field is
`\angle =90-\theta`

And flux is given by

`\phi =\vec{E}\cdot \vec{A}`

`\phi =E\cdot A\cos (90-\theta )`

`2* 10^6=6* 10^5* 7* \cos (90-\theta )`

`\sin \theta =0.476`

thus
`\theta =28.42^(\circ)`