# Calculating electric flux with a constant electric field

1. Aug 2, 2011

### NewtonianAlch

1. The problem statement, all variables and given/known data

A vertical electric field of magnitude 2 X 10^4 N/C exists above the Earth's surface on a day when a thunderstorm is brewing. A car with a rectangular size of 3.00 m by 1.50 m is traveling along a dry gravel roadway sloping downward at 7.0°. Determine the electric flux through the bottom of the car.

2. Relevant equations

Eflux = E*(A cos theta)

3. The attempt at a solution

(2*10^4)*(3*1.5)*(cos 7) = 89,329 Nm^2/C = 8.93*10^4 Nm^2/C

Is this correct? It seems pretty straightforward to me, but I can't help think that there is some trick in the question that I'm not seeing.

Flux through a surface perpendicular to the electric field obviously has cos 90 = 0 as the angle angle associated with it so it would just be Eflux = E*A, but there is an angle here, so it would simply be cos 7. Does the thunderstorm brewing have any bearing on the equations affecting the answer?

Last edited: Aug 2, 2011
2. Aug 2, 2011

### Disconnected

It seems the thunderstorm that is brewing is the source of the E field, and is just a little back story.

Have a quick thought about it. The X flux is the amount of X going through a surface area, correct?

If the field is uniform and verticle, will there be a difference between the flux through a horizontal surface A and a surface B, if surface B's horizontal componant of area (oh gawd did I just slaughter that phrase, but I hope you get my point) is the same as surface A?

For example, if there is a sheet area 1m by 3m held horizontally in a rain storm (assuming uniform verticle rain) will it's dry patch be the same area as a 1m by 5m sheet held at 53* (or whatever the angle in a 3,4,5 triangle is)?

So you are really just finding the surface area seen by the uniform field, then multiplying that by the strength of that field. It is very simple.