 #1
 54
 0
Homework Statement
A closed surface with dimensions a = b =
0.254 m and c = 0.4064 m is located as in
the ﬁgure. The electric ﬁeld throughout the
region is nonuniform and given by ~E = (α +
β x^2)ˆı where x is in meters, α = 4 N/C, and
β = 6 N/(C m2).
Picture of object attached.
What is the magnitude of the net charge
enclosed by the surface?
Answer in units of C.
Homework Equations
Gauss's Law: Flux = integral(E dA) = Q/permittivity constant
permittivity constant(epsilon naught) = 8.85E12
The Attempt at a Solution
I know that the object doesn't not have any flux through any of the sides except the left and right sides(those parallel to the electric field). So I thought to find net flux and multiply that by the permittivity constant and find the net charge enclosed.
Flux1 = (α +β (a+c)^2)*(a*b) = Electric field * area of the side(right)
Flux2 = (α +β (a)^2) * (a*b) = Electric field * area of the other side(left)
Then:
Flux1 + Flux2 = 0.709925351759
Therefore I multiply by epsilon naught:
net flux * epsilon naught = 6.28283939307E12
However this is wrong, I believe I am messing up because of the weird way they are giving the electric field as, or possibly my entire calculations or process?
Attachments

26.1 KB Views: 673