Answer with Explanation:
We are given that
Density=[tex]\rho l[/tex]
A(4m,2m,4m) and B(0,0,0)
y=1 m
a. Linear charge density=[tex]\frac{\rho l}{l}=\rho C/m[/tex]
Let a point P (0,1,4) on the line of charge and point Q (0,1,0)
Therefore,
Distance AP=[tex]\sqrt{(4-0)^2+(2-1)^2+(4-4)^2}=\sqrt{17}[/tex]
Distance,BQ=[tex]\sqrt{(0-0)^2+(1-0)^2+(0-0)^2}=1[/tex]
Electric field for infinitely long line
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}[/tex]
Therefore, potential
[tex]V_{BA}=-\int_{a}^{b}E\cdot dl[/tex]
[tex]V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}\hat{r}\cdot \hat{r} dr[/tex]
[tex]V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}dr[/tex]
[tex]V_{BA}=-\frac{\rho}{2\pi \epsilon_0}[\ln r]^{1}_{\sqrt{17}}[/tex]
[tex]V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln 1-ln(\sqrt{17})=\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})[/tex]
[tex]V_{BA}=V_B-V_A[/tex]
[tex]V_{AB}=V_A-V_B=-V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})[/tex]
b.Electric field at point B
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}[/tex]
Unit vector r=[tex]-\hat{j}[/tex]
Therefore,
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{-j}[/tex]
