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The electric field E = (5.4, 2.7, 8.1) N/C passes through a square surface whose area vector is A = (1.7, −3.4, 0.55) m2. What is the electric flux ΦE through the surface?
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If the square surface has no charge then the electric flux will also be 0 \[\phi =\]|dw:1316653397896:dw|
I think Mustaafa isn't quite right, what he stands is true (and it is Gauss' Law) only for closed surfaces, i.e., the Flux \[\Phi _{E}=\frac{q}{\epsilon_{0}}\], where \[\ q \] is the total charge enclosed by the CLOSED surface. In this case we have an open surface, so you should calculate the flux through it using: \[\ \Phi_{E} = E * A \], where * stands for the Dot product. Good Luck ;)
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