# Flux above a flat horizontal surface

1. Sep 17, 2014

### sea95

So my question is, am I missing any key thing in answering the question below?

Question: An electric flux of 149 N.m2/C passes through a flat horizontal surface that has an area of 0.67 m2. The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15degrees above the horizontal?

So I've tried E=I/(Acos(90-θ)
I=flux, and A= area, 0.67

E= 149/(0.67cos75)

And I have gotten...

=859.24 (this seemed too large)

=204.98 (I didn't use parenthesis, and too low)

=57.55 (this was just an odd one my calculator gave me)

So, I believe that it should fall somewhere in the 400-650ish range but please help, I have no idea of what to do with my calculator's answers

2. Sep 21, 2014

### Simon Bridge

... what leads you to that belief?

If the angle is 0, then E=I/A = 222.39N/C - which would be a minimum, right?
As the angle gets shallower, then you need a bigger E field to get the same flux.

Your belief is that 15deg should give you 2-3 times the minimum field required for the flux?
What is $1/\sin(15^\circ)$ ? You don't have to trust your calculator if you don't want to - figure the trig by hand.