Electric field strength (should be easy)

[FrogPad]

What is the electric field strength in a region where the flux through a: 1.0cm \times 1.0cm flat surface is: 65N\frac{m^2}{C}, if the field is uniform and the surface is at right angles to the field?

[]
[]----> E
[]

Since the field points at a right angle to the surface we have a norm of 1. (cos(90) = 1).

So we have:

<br /> \phi = 65N\frac{m^2}{C}<br />
<br /> A = (1.0cm)^2 = (0.01m)^2 = 0.0001m^2<br />
<br /> E= unknown<br />

Therefore:

<br /> \phi = E*A<br />

<br /> 65N\frac{m^2}{C}=E (0.0001)<br />
<br /> E=\frac{65}{0.0001}\frac{Nm^2}{C}<br />

Thus:
<br /> E=650000N\frac{m^2}{C}<br />

Did I do this math correctly? I just feel like I am doing something wrong here. Thanks

 

[MathStudent]

Your answer is correct, but let me just point something out.
The norm is a unit vector that is perpendicular to the surface, so in this case its parallel to the electric field so the angle (theta) between the norm and the E is 0. cos(0) = 1,
cos(90) = 0 ( I think you knew this but just made careless/typographical error )

 

[FrogPad]

hehe, so careful with latex... not so careful with plain text.

I just started this physics course, so all the reinforcement (even mistakes) is awesome... thanks for the double check :)