Question

The figure shows a closed Gaussian surface in the shape of cube of edge length 2.1 m, with one corner at x_1 = 4.0m, y_1 = 4.1 m. The cube lies in a region where the electric filed vector is given by E = -2.4 I - 3.6y^2j + 3.4 k N/c, with y in meter. What is the net charge contained by the cube?

Answer 1

$\dpi{100} \small \triangledown.E = -7.2(N/Cm)$

From the divergence theorem we have

$\dpi{100} \small \int \int E.ds=\int \int \int \triangledown E.dV$

the flux can be calculated by integrating -7.2 y over the volume of the cube. That equals $\dpi{100} \small Q/\epsilon _{o}$

. $\dpi{100} \small Q/\epsilon _{o} = \int dx\int dy\int dz(-7.2y)$

$\dpi{100} \small = \int dx\int dz \int dy(-7.2y)$

right side y of the cube y = 4.1 and left side of the cube y = 4.1-2.1 = 2.0

$\dpi{100} \small Q/\epsilon _{o}= 2.1 \times 2.1 \times \left [ -3.6y^{2}\right ]^{2.0}_{4.1}$

= -3.6 x 2.1 x 2.1 x ((4.1)² - (2.0)²) N/Cm²

$\dpi{100} \small Q/\epsilon _{o} = -203.37Nm^{2}/C$  

Q = (8.854x10^-12C²/Nm²) x (-203.37 Nm²/C)

the net charge contained by the cube

Q = -1.8x10^-9C