91影视

Using the gauss flux theorem, we can get the net flux through the surfaces. Now, using the same concept, we can get the net charge contained in the cube.

Formula:

The electric flux passing through the surface,

$\mathrm{蠒}=\mathrm{E}.\mathrm{A}=\frac{\mathrm{q}}{{\mathrm{蔚}}_0}$ (1)

None of the constant terms will result in a nonzero contribution to the flux, so we focus on the x-dependent term only. In Si units, we have

${\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}=3\mathrm{x}\hat{\mathrm{i}}$

The face of the cube located at x = 0 (in the y-z plane) has area $\mathrm{A}=4{\mathrm{m}}^2$(and it 鈥渇aces鈥� the +i direction) and has a 鈥渃ontribution鈥� to the flux that is given using equation (1) such that,

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}}^{\mathrm{A}}=\left(\right(3.00\left)\right(0)\mathrm{N}/\mathrm{C}(4{\mathrm{m}}^2) \\ =0 \end{gathered}$$

The face of the cube located at x = -2m has the same area A (and this one 鈥渇aces鈥� the 鈥搃 direction) and a contribution to the flux that is given using equation (i) as:

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}}^{\mathrm{A}}=-\left(\right(3.00\left)\right(-2)\mathrm{N}/\mathrm{C}(4{\mathrm{m}}^2) \\ =24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Thus, the net flux is given by:

$$\begin{gathered} \mathrm{蠒}=0\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}+24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

According to Gauss鈥� law, the net enclosed charge by the cube is given using equation (1) as:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}=(8.85脳{10}^{-12}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2)(24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}) \\ =2.13脳{10}^{-10}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $2.13脳{10}^{-10}\mathrm{C}$.