1. calculate d in rectangular coordinates at point p(2, -3,6) produced by:\n a. a point charge $q_a = 55 mc$…

Explanation:

Step1: Electric - field due to point charge

The electric - displacement vector $\mathbf{D}$ due to a point charge $Q$ at a distance $\mathbf{r}$ is given by $\mathbf{D}=\frac{Q}{4\pi r^{2}}\hat{\mathbf{r}}$. First, find the position vector $\mathbf{r}=\mathbf{r}_P-\mathbf{r}_Q=(2 - (- 2))\hat{\mathbf{i}}+(-3 - 3)\hat{\mathbf{j}}+(6-(-6))\hat{\mathbf{k}} = 4\hat{\mathbf{i}}-6\hat{\mathbf{j}}+12\hat{\mathbf{k}}$. The magnitude $r=\sqrt{4^{2}+(-6)^{2}+12^{2}}=\sqrt{16 + 36+144}=\sqrt{196}=14$. Then $\mathbf{D}_a=\frac{Q_A}{4\pi r^{2}}\frac{\mathbf{r}}{r}$, where $Q_A = 55\times10^{-3}\ C$. So $\mathbf{D}_a=\frac{55\times10^{-3}}{4\pi\times14^{2}}\frac{4\hat{\mathbf{i}}-6\hat{\mathbf{j}}+12\hat{\mathbf{k}}}{14}=\frac{55\times10^{-3}(4\hat{\mathbf{i}}-6\hat{\mathbf{j}}+12\hat{\mathbf{k}})}{4\pi\times14^{3}}\ C/m^{2}$.

Step2: Electric - field due to line charge

The electric - displacement vector $\mathbf{D}$ due to an infinite line charge with linear charge density $\rho_L$ at a perpendicular distance $\rho$ from the line charge (in cylindrical coordinates, but we can convert to rectangular) is $\mathbf{D}=\frac{\rho_L}{2\pi\rho}\hat{\rho}$. In rectangular coordinates, for a line charge on the $x$ - axis, $\rho=\sqrt{(-3)^{2}+6^{2}}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}$. The unit vector $\hat{\rho}=\frac{-3\hat{\mathbf{j}}+6\hat{\mathbf{k}}}{3\sqrt{5}}$. Then $\mathbf{D}_b=\frac{\rho_L}{2\pi\rho}\hat{\rho}$, with $\rho_L = 20\times10^{-3}\ C/m$. So $\mathbf{D}_b=\frac{20\times10^{-3}}{2\pi\times3\sqrt{5}}\frac{-3\hat{\mathbf{j}}+6\hat{\mathbf{k}}}{3\sqrt{5}}\ C/m^{2}=\frac{20\times10^{-3}(-3\hat{\mathbf{j}}+6\hat{\mathbf{k}})}{2\pi\times45}\ C/m^{2}$.

Step3: Electric - field due to surface charge

The electric - displacement vector $\mathbf{D}$ due to an infinite surface charge with surface charge density $\rho_S$ is $\mathbf{D}=\rho_S\hat{\mathbf{n}}$. The plane is $z=-5\ m$, and the point $P(2,-3,6)$ is above the plane. The unit - normal vector $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ (since the surface is in the $z = - 5$ plane and we are above it). So $\mathbf{D}_c=\rho_S\hat{\mathbf{k}}$, with $\rho_S = 120\times10^{-6}\ C/m^{2}$, then $\mathbf{D}_c=120\times10^{-6}\hat{\mathbf{k}}\ C/m^{2}$.

Step4: Total electric - displacement vector

$\mathbf{D}=\mathbf{D}_a+\mathbf{D}_b+\mathbf{D}_c$. [ \begin{align*} \mathbf{D}&=\frac{55\times10^{-3}(4\hat{\mathbf{i}}-6\hat{\mathbf{j}}+12\hat{\mathbf{k}})}{4\pi\times14^{3}}+\frac{20\times10^{-3}(-3\hat{\mathbf{j}}+6\hat{\mathbf{k}})}{2\pi\times45}+120\times10^{-6}\hat{\mathbf{k}}\ \end{align*} ] First, calculate the coefficients: [ \begin{align*} \frac{55\times10^{-3}\times4}{4\pi\times14^{3}}&=\frac{220\times10^{-3}}{4\pi\times2744}\approx\frac{220\times10^{-3}}{34486.56}\approx6.38\times10^{-6}\ \frac{- 55\times10^{-3}\times6}{4\pi\times14^{3}}&=\frac{-330\times10^{-3}}{4\pi\times2744}\approx - 9.57\times10^{-6}\ \frac{55\times10^{-3}\times12}{4\pi\times14^{3}}&=\frac{660\times10^{-3}}{4\pi\times2744}\approx19.14\times10^{-6}\ \frac{-20\times10^{-3}\times3}{2\pi\times45}&=\frac{-60\times10^{-3}}{2\pi\times45}\approx - 2.12\times10^{-4}\ \frac{20\times10^{-3}\times6}{2\pi\times45}&=\frac{120\times10^{-3}}{2\pi\times45}\approx4.24\times10^{-4} \end{align*} ] [ \begin{align*} \mathbf{D}&=(6.38\times10^{-6}\hat{\mathbf{i}}+(-9.57\times10^{-6}-2.12\times10^{-4})\hat{\mathbf{j}}+(19.14\times10^{-6}+4.24\times10^{-4}+120\times10^{-6})\hat{\mathbf{k}})\ &=(6.38\times10^{-6}\hat{\mathbf{i}}+(-2.22\times10^{-4})\hat{\mathbf{j}}+(5.63\times10^{-4})\hat{\mathbf{k}})\ C/m^{2} \end{align*} ]

Answer:

$\mathbf{D}=(6.38\times10^{-6}\hat{\mathbf{i}}-2.22\times10^{-4}\hat{\mathbf{j}}+5.63\times10^{-4}\hat{\mathbf{k}})\ C/m^{2}$