Chapter 6, Problem 2P

(a)

To determine

Calculate the electric field E at $P\left(1,60°,30°\right)$.

Answer to Problem 2P

The electric field E at $P\left(1,60°,30°\right)$ is $\underset{\_}{5a_r+4.33a_{\theta }-5a_{\varphi }\frac{\text{V}}{\text{m}}}$.

Explanation of Solution

Calculation:

Calculate the electric field.

$\begin{array}{c}-E=\nabla V\\ =\nabla \left(\frac{10\cos \theta \sin \varphi }{r^2}\right)\\ =\left(-\frac{20}{r^3}\cos \theta \sin \varphi a_r-\frac{10}{r^3}\sin \theta \sin \varphi a_{\theta }+\frac{1}{r\sin \theta }\frac{10}{r^2}\cos \theta \cos \varphi a_{\varphi }\right)\frac{\text{V}}{\text{m}}\end{array}$

$E=\left(\frac{20}{r^3}\cos \theta \sin \varphi a_r+\frac{10}{r^3}\sin \theta \sin \varphi a_{\theta }-\frac{1}{r\sin \theta }\frac{10}{r^2}\cos \theta \cos \varphi a_{\varphi }\right)\frac{\text{V}}{\text{m}}$        (1)

Consider that the given point $P\left(1,60°,30°\right)$.

Substitute 1 for r, $60°$ for $\theta$, and $30°$ for $\varphi$ in Equation (1).

$\begin{array}{c}E=\left(\frac{20}{1^3}\cos 60°\sin 30°a_r+\frac{10}{1^3}\sin 60°\sin 30°a_{\theta }-\frac{1}{1\sin 60°}\frac{10}{1^2}\cos 60°\cos 30°a_{\varphi }\right)\frac{\text{V}}{\text{m}}\\ =5a_r+4.33a_{\theta }-5a_{\varphi }\frac{\text{V}}{\text{m}}\end{array}$

Conclusion:

Thus, the electric field E at $P\left(1,60°,30°\right)$ is $\underset{\_}{5a_r+4.33a_{\theta }-5a_{\varphi }\frac{\text{V}}{\text{m}}}$.

(b)

To determine

Find the volume charge density ${\rho }_v$ at $P\left(1,60°,30°\right)$.

Answer to Problem 2P

The volume charge density ${\rho }_v$ is $\underset{\_}{29.47\frac{\text{pC}}{{\text{m}}^3}}$.

Explanation of Solution

Calculation:

Consider the expression for the Poisson’s equation.

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

Re-arrange the equation to find the volume charge density.

${\rho }_v=-\epsilon {\nabla }^{2}V{\text{C/m}}^3$        (2)

Calculate the value of ${\nabla }^{2}V$.

$\begin{array}{c}{\nabla }^{2}V={\nabla }^2\left(\frac{10\cos \theta \sin \varphi }{r^2}\right)\\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(\frac{-20\cos \theta \sin \varphi }{r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\frac{-10{\sin }^2\theta \sin \varphi }{r^2}\right)-\frac{1}{r^2{\sin }^2\theta }\frac{10\cos \theta \sin \varphi }{r^2}\\ =\frac{20\cos \theta \sin \varphi }{r^4}-\frac{20\sin \theta \cos \theta \sin \varphi }{r^4\sin \theta }-\frac{10\cos \theta \sin \varphi }{r^4{\sin }^2\theta }\\ =-\frac{10\cos \theta \sin \varphi }{r^4{\sin }^2\theta }\end{array}$

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

$\begin{array}{c}{\rho }_v=-{\epsilon }_o\left(-\frac{10\cos \theta \sin \varphi }{r^4{\sin }^2\theta }\right)\frac{\text{C}}{{\text{m}}^3}\\ =\frac{10{\epsilon }_o\cos \theta \sin \varphi }{r^4{\sin }^2\theta }\frac{\text{C}}{{\text{m}}^3}\end{array}$

Substitute $\frac{{10}^{-9}}{36\pi }\text{F}\cdot {\text{m}}^{-1}$ for ${\epsilon }_o$, 1 for r, $60°$ for $\theta$, and $30°$ for $\varphi$.

$\begin{array}{c}{\rho }_v=\frac{10\left(\frac{{10}^{-9}}{36\pi }\right)\cos 60°\sin 30°}{1^4{\sin }^{2}60°}\frac{\text{C}}{{\text{m}}^3}\\ =29.47\times {10}^{-12}\frac{\text{C}}{{\text{m}}^3}\\ =29.47\frac{\text{pC}}{{\text{m}}^3}\end{array}$

Conclusion:

Thus, the volume charge density ${\rho }_v$ is $\underset{\_}{29.47\frac{\text{pC}}{{\text{m}}^3}}$.