URGENT!!!!! The box-like Gaussian surface of the figure encloses a net charge of
ID: 1435132 • Letter: U
Question
URGENT!!!!!
The box-like Gaussian surface of the figure encloses a net charge of 24.0?0 C and lies in an electric field given by ModifyingAbove Upper E With right-arrow equals left-bracket left-parenthesis 9.00 plus 3.20 x right-parenthesis ModifyingAbove i With caret minus 4.50 ModifyingAbove j With caret plus b z ModifyingAbove k With caret right-bracket N/C with x and z in meters and b a constant. The bottom face is in the xz plane; the top face is in the horizontal plane passing through y2 = 1.00 m. For x1 = 1.20 m, x2 = 4.30 m, z1 = 1.20 m, and z2 = 2.80 m, what is b?
Explanation / Answer
E = [(9+3.20x)i -4.50 j + bzk]
phi = integral(E.dA)
The net flux through the two faces parallel to the yz plane is
phi_yz = double integral Ex(x=x2) - Ex(x=x1)] dydz = from y1=0 to y2 = 1 dy from z1 = 1,2 to z2 = 2.8 integral dz[ 9 +3.2(4.3) - 9-3.2(1.2)]
= 9.92 * ( 1) * (1.6) = 15.872
Similarly, the net flux through the two faces parallel to the xz plane is
phi_xz = 0
the net flux through the two faces parallel to the xy plane is
phi_xy = double integral [ Ez(z=z2)-Ez(Z=z1)]dxdy
from x1 = 1.2 to x2 = 4.3 integral dx integral dy from y1 = 0 to y2 = 1 dy (2.8b - 1.2b)
phiu_xy = 4.96b
Qenc = eo*phi = eo*(15.872 + 4.96b) = 24 eo
b = 1.6387 N/C.m
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