# Electric field strength (should be easy)

1. Feb 2, 2005

### 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:

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

Therefore:

$$\phi = E*A$$

$$65N\frac{m^2}{C}=E (0.0001)$$
$$E=\frac{65}{0.0001}\frac{Nm^2}{C}$$

Thus:
$$E=650000N\frac{m^2}{C}$$

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

2. Feb 2, 2005

### 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 )

3. Feb 2, 2005

### 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 :)

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