Chapter 6, Problem 1P

(a)

To determine

Calculate the electric field at $P\left(-3,1,2\right)$.

Answer to Problem 1P

The electric field at $P\left(-3,1,2\right)$ is $\underset{\_}{-270a_x+540a_y+135a_z\frac{\text{V}}{\text{m}}}$.

Explanation of Solution

Calculation:

Write the expression for the electric field.

$\begin{array}{c}E=-\nabla V\\ =-\left(\frac{\partial V}{\partial x}{a}_x+\frac{\partial V}{\partial y}{a}_y+\frac{\partial V}{\partial z}{a}_z\right)\frac{\text{V}}{\text{m}}\end{array}$

Substitute $5x^3{y}^{2}z$ for V.

$E=-\left(\frac{\partial \left(5x^3{y}^{2}z\right)}{\partial x}{a}_x+\frac{\partial \left(5x^3{y}^{2}z\right)}{\partial y}{a}_y+\frac{\partial \left(5x^3{y}^{2}z\right)}{\partial z}{a}_z\right)\frac{\text{V}}{\text{m}}$

$E=-\left(15x^2{y}^{2}z{a}_x+10x^{3}yz{a}_y+5x^3{y}^2{a}_z\right)\frac{\text{V}}{\text{m}}$        (1)

Consider that the given $P\left(-3,1,2\right)$.

Substitute $-3$ for x, 1 for y, and 2 for z in Equation (1).

$\begin{array}{c}E=-\left(15{\left(-3\right)}^2{\left(1\right)}^2\left(2\right)a_x+10{\left(-3\right)}^3\left(1\right)\left(2\right)a_y+5{\left(-3\right)}^3{\left(1\right)}^2{a}_z\right)\frac{\text{V}}{\text{m}}\\ =-270a_x+540a_y+135a_z\frac{\text{V}}{\text{m}}\end{array}$

Conclusion:

Thus, the electric field at $P\left(-3,1,2\right)$ is $\underset{\_}{-270a_x+540a_y+135a_z\frac{\text{V}}{\text{m}}}$.

(b)

To determine

Find the volume charge density ${\rho }_v$ at $P\left(-3,1,2\right)$.

Answer to Problem 1P

The volume charge density ${\rho }_v$ is $\underset{\_}{14.324\frac{\text{nC}}{{\text{m}}^3}}$.

Explanation of Solution

Calculation:

Consider the expression for the volume charge density.

${\rho }_v=-\epsilon {\nabla }^{2}V$

${\rho }_v=-\epsilon \left(\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}\right)$        (2)

Here,

${\nabla }^{2}V=\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}$

Substitute $5x^3{y}^{2}z$ for V and $2.25{\epsilon }_o$ for $\epsilon$ in Equation (2).

$\begin{array}{c}{\rho }_v=-2.25{\epsilon }_o\left(\frac{{\partial }^2\left(5x^3{y}^{2}z\right)}{\partial x^2}+\frac{{\partial }^2\left(5x^3{y}^{2}z\right)}{\partial y^2}+\frac{{\partial }^2\left(5x^3{y}^{2}z\right)}{\partial z^2}\right)\\ =-2.25{\epsilon }_o\left[\frac{\partial }{\partial x}\left(15x^2{y}^{2}z\right)+\frac{\partial }{\partial y}\left(10x^{3}yz\right)+\frac{\partial }{\partial z}\left(5x^3{y}^2\right)\right]\\ =-2.25{\epsilon }_o\left[30x{y}^{2}z+10x^{3}z\right]\end{array}$

Substitute $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, $-3$ for x, 1 for y, and 2 for z.

$\begin{array}{c}{\rho }_v=-2.25\left(\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}\right)\left[30\left(-3\right){\left(1\right)}^2\left(2\right)+10{\left(-3\right)}^3\left(2\right)\right]\frac{\text{V}}{{\text{m}}^{\text{2}}}\\ =-2.25\left(\frac{{10}^{-9}}{36\pi }\frac{\text{C}}{\text{V}}\cdot {\text{m}}^{-1}\right)\left[30\left(-3\right){\left(1\right)}^2\left(2\right)+10{\left(-3\right)}^3\left(2\right)\right]\frac{\text{V}}{{\text{m}}^{\text{2}}}\\ =14.324\frac{\text{nC}}{{\text{m}}^3}\left\{\because \text{C}=\text{F}\cdot \text{V}\right\}\end{array}$

Conclusion:

Thus, the volume charge density ${\rho }_v$ is $\underset{\_}{14.324\frac{\text{nC}}{{\text{m}}^3}}$.