Question 4.7: Determine D at (4, 0, 3) if there is a point charge -5π mC a...

Let D=D_{Q}+D_{L}, where D_{Q} and D_{L} are flux densities due to the point charge and line charge, respectively, as shown in Figure 4.11:

D_{Q}=\varepsilon _{o}E=\frac{Q}{4\pi R^{2}}a_{R}=\frac{Q\left(r-r^{'}\right) }{4\pi\left|r-r^{'}\right|^{3} }

where

r-r^{'}=\left(4,0,3\right)-\left(4,0,0\right)=\left(0,0,3\right)

Hence

D_{Q}=\frac{-5\pi\cdot 10^{-3}\left(0,0,3\right) }{4\pi\left|\left(0,0,3\right)\right|^{3}}=-0.139a_{z}mC/m^{2}

Also

D_{L}=\frac{\rho _{L}}{2\pi\rho }a_{\rho }

In this case

a_{\rho }=\frac{\left(4,0,3\right)-\left(0,0,0\right)}{\left|\left(4,0,3\right)-\left(0,0,0\right) \right| }=\frac{\left(4,0,3\right)}{5}

\rho =\left|\left(4,0,3\right)-\left(0,0,0\right) \right|=5

Hence

D_{L}=\frac{3\pi}{2\pi\left(25\right) }\left(4a_{x}+3a_{z}\right)=0.24a_{x}+0.18a_{z}mC/m^{2}

Thus

D=D_{Q}+D_{L}=240a_{x}+41.1a_{z}\mu C/m^{2}