Chapter 4, Problem 1P

A cube 2 m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density is given by ρ_v = xy²e^−2z (mC/m³).

To determine

The total charge contained in the cube for the given conditions.

Answer to Problem 1P

The total charge contained in the cube for the given conditions is $2.62\text{mC}$.

Explanation of Solution

Given data:

The side of the cube is $2\text{m}$.

The volume charge density is ${\rho }_{\text{v}}=x{y}^2{e}^{-2z}\left(\text{mC}/{\text{m}}^{\text{3}}\right)$.

Calculation:

The required diagram for the given condition is drawn as shown in Figure 1.

Write the standard expression of charge for volume charge density.

$Q={\displaystyle \underset{\upsilon }{\int }{\rho }_{\text{v}}d\upsilon }$                                                                                                     (1)

Here,

$Q$ is the charge contained in the cube.

$d\upsilon$ is the differential volume.

Differential volume in Cartesian coordinate system is,

$d\upsilon =dx\times dy\times dz$

The given cube is resting on one of its corner at origin and in the first quadrant.

Hence, range of $x\text{-}$ coordinate is $0\le x\le 2\text{m}$.

Range of $y\text{-}$ coordinate is $0\le y\le 2\text{m}$.

Range of $z\text{-}$ coordinate is $0\le z\le 2\text{m}$.

Substituting $x{y}^2{e}^{-2z}$ for ${\rho }_{\text{v}}$, $dx\times dy\times dz$ for $d\upsilon$, $0$ for $x_1$, $0$ for $y_1$, $0$ for $z_1$, $2$ for $x_2$, $2$ for $y_2$, $2$ for $z_2$ in equation (1).

$\begin{array}{c}Q={\displaystyle \underset{x_1=0}{\overset{x_2=2}{\int }}{\displaystyle \underset{y_1=0}{\overset{y_2=2}{\int }}{\displaystyle \underset{z_1=0}{\overset{z_2=2}{\int }}x{y}^2{e}^{-2z}\times dx\times dy\times dz}}}\\ ={\displaystyle \underset{x_1=0}{\overset{x_2=2}{\int }}xdx}\times {\displaystyle \underset{y_1=0}{\overset{y_2=2}{\int }}{y}^{2}dy}\times {\displaystyle \underset{z_1=0}{\overset{z_2=2}{\int }}{e}^{-2z}dz}\\ ={\left(\frac{x^2}{2}\right)}_0^2\times {\left(\frac{y^3}{3}\right)}_0^2\times {\left(\frac{e^{-2z}}{-2}\right)}_0^2\end{array}$

Solve the above expression.

$\begin{array}{c}Q=\left(\frac{2^2}{2}\right)\times \left(\frac{2^3}{3}\right)\times \left(\frac{e^{-4}-e^0}{-2}\right)\\ =\left(2\right)\left(\frac{8}{3}\right)\times \left(0.491\right)\\ =2.62\text{mC}\end{array}$

Conclusion:

Therefore, the total charge contained in the cube is $Q=2.62\text{mC}$.