Question

An infinitely long wire is kept along :- axis from $z=-\infty$ to $z=+\infty ,$ having uniform linear charge density $\frac{10}{9}nC/m$ The electric field at point (6 cm, 8 cm, 10 cm) will be

• A. $(120\hat{i}+160\hat{j}+200\hat{k})V/m$
• B. $200\hat{k}V/m$
• C. $(160\hat{i}+120\hat{j})V/m$
• D. $(120\hat{i}+160\hat{j})V/m$

Answer 1

The correct option is A $(120\hat{i}+160\hat{j}+200\hat{k})V/m$

$z=-\infty$ to $z=+\infty$

density $=\frac{10}{9}nc/m$ the electric field at the electrostatic force on $q$ is the vector super position of the following three forces

${\overline{F}}_1=\frac{K_{q}Q}{L^2}x$

${\overline{F}}_2=-\frac{K_{q}Q}{L^2}\overline{y}$

${\overline{F}}_3=\frac{2K_{q}Q}{{\left(\sqrt{2}L\right)}^2}(\frac{1}{\sqrt{2}}\hat{x}+\frac{1}{\sqrt{2}}\hat{y})$

and the net force is thus

${\overline{F}}_0=\frac{K_{q}Q}{L^2}(1-\frac{1}{\sqrt{2}})(\hat{x}-q)$

As we learnt

If $\rho$ lies outside

$E_{out}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}{V}_{out}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$

$E_{out}=\frac{{\rho R}^2}{3{ϵ}_0{r}^2}{V}_{out}=\frac{{\rho R}^3}{3{ϵ}_{0}r}$

Total charge on sphere

$q={\int }_0^R\rho dv={\int }_0^R{\rho }_0(1-\frac{r}{R})4\pi r^{2}dr$

$q={\rho }_{0}4\pi [{\int }_0^R{r}^{2}dr-{\int }_0^R\frac{r^3}{R}dr]$

$q={\rho }_{0}4\pi (\frac{R^3}{3}-\frac{R^3}{4})$

$={\rho }_{0}4\pi \frac{R^3}{12}$

$q=\frac{{\rho }_0\pi R^3}{3}$

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{{\rho }_0\pi r^3}{3r^2}=\frac{{\rho }_0{R}^3}{12{ϵ}_0{r}^2}$

The magnitude is given by

$\left|{\overline{F}}_{tot}\right|=\frac{K_{q}Q}{L^2}(\sqrt{2}-1)$

$(120\hat{i}+160\hat{j}+200\hat{k})v/m$