102 Week 2: Continuous Charge and Gauss’s Law
Advanced Problem 12.
xy∆x∆y
∆z(x 0 ,y 0 ,z 0 )zConsider asmallgaussian surface in the shape of a cube with faces parallel to thexy,xz, andyz
planes sitting in region where there is a continuous electric field. Let the corner nearest the origin be
located at~r 0 = (x 0 , y 0 , z 0 ) and the cube edge lengths be ∆x= ∆y= ∆zin the directions parallel
to the different axes.
Since the electric field is continuous, each component of the field canbe expanded in a Taylor
series:
E~(~r 0 + ∆~r) =
(
Ex(~r 0 ) + ∆x∂Ex
∂x∣∣
∣∣
~r 0+ ∆y∂Ex
∂y∣∣
∣∣
~r 0+
∆z∂Ex
∂z∣∣
∣∣
~r 0+...
)
xˆ+
(
Ey(~r 0 ) + ∆x∂Ey
∂x∣∣
∣∣
~r 0+ ∆y∂Ey
∂y∣∣
∣∣
~r 0+
∆z∂Ey
∂z∣∣
∣∣
~r 0+...
)
yˆ+
(
Ez(~r 0 ) + ∆x∂Ez
∂x∣∣
∣∣
~r 0+ ∆y∂Ez
∂y∣∣
∣∣
~r 0+
∆z∂Ez
∂z∣∣
∣∣
~r 0+...
)
ˆz+(122)where we only keep/show first order terms.
Noting that ∆A= ∆x∆y= ∆x∆z= ∆z∆y(depending on the side) and that ∆V= ∆x∆y∆z,
show that the net electric fluxoutof this box is:
∑sidesE~·ˆn∆A=φnet=(
∂Ex
∂x+
∂Ey
∂y+
∂Ez
∂z