Flux and Gauss' Law

PHY110W ELECTRICITY SUPPLEMENTARY GAUSS’ LAW
Type 2. A cubic Gaussian surface is bounded by the planes at x = 0.40 m,
y = 0.60 m and z = 0.40 m as shown in the sketch. The electric field in this
region is non-uniform and is given by E = (2.00 + 3.00x 2 )î N/C. Calculate:
0.6
y
(a)
the flux through each face of the cube;
0.2
(b) the net charge enclosed by the cube.
Solution: (a) This time we must apply Φ = E ⋅dA
over each of the six faces –
∫
z 0.4
0.4
which is actually not as much work as it seems, because the field has no components in the y
or z directions, so there will be no flux through the top, bottom, front and back faces:
Ie Φ top = Φ bottom = Φ front = Φ back = 0 N⋅m 2 /C
We’ve already calculated the flux through the right hand face (in Type 1), so we only have to
follow the same procedure for the left face, where x = 0 m.
Here the electric field is a constant (2.00 + 3.00 × 0 2 ) = 2.00 N/C and again lies in the positive
x direction. This time, however, the area vector points in the negative x direction (because it
always points out of the Gaussian surface), so cosθ = –1, and we get
Φ
left
= E∫ dA =−E A = –2.0 × 0.40 2 = –0,32 N⋅m 2 /C
(b) Using Gauss’ Law, qenclosed
= ε
0Φ (where Φ is the total flux through all six faces)…
∴q enclosed = 8.85 × 10 -12 × (0 + 0 + 0 + 0 + 0.397 + (–0.32)) = +6.80 × 10 -13 C
x
4