What is the electric field strength in a region where the flux through a: [tex]1.0cm \times 1.0cm[/tex] flat surface is: [tex]65N\frac{m^2}{C}[/tex], if the field is uniform and the surface is at right angles to the field?(adsbygoogle = window.adsbygoogle || []).push({});

[]

[]----> E

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

So we have:

[tex]

\phi = 65N\frac{m^2}{C}

[/tex]

[tex]

A = (1.0cm)^2 = (0.01m)^2 = 0.0001m^2

[/tex]

[tex]

E= unknown

[/tex]

Therefore:

[tex]

\phi = E*A

[/tex]

[tex]

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

[/tex]

[tex]

E=\frac{65}{0.0001}\frac{Nm^2}{C}

[/tex]

Thus:

[tex]

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

[/tex]

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

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# Homework Help: Electric field strength (should be easy)

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