- #1
FrogPad
- 810
- 0
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?
[]
[]----> E
[]
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
[]
[]----> E
[]
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