Question

An infinitely long wire is kept along $z$ -axis from $z=-\in fty$ to $z=+\infty$ , having uniform linear charge density $\frac{10}{9}nC{m}^{-1}$ The electric field $\overrightarrow{E}$ at point $\left(6cm,8cm,10cm\right)$ will be:

• A. $(160\hat{i}+120\hat{j}+200\hat{k})N{C}^{-1}$
• B. $(200\hat{k})N{C}^{-1}$
• C. $(160\hat{i}+120\hat{j})N{C}^{-1}$
• D. $(120\hat{i}+160\hat{j})N{C}^{-1}$

Answer 1

Answer: (D)

$E=\frac{2k\lambda }{d}=\frac{2\times 9\times 10^9\times \left(\frac{10}{9}\right)\times 10^{-9}}{10\times 10^{-2}}=200\mathrm{V}\mathrm{m}{\mathrm{m}}^{-1}$
As field is radial out, there is no component of it in $Z$ direction.
$\overrightarrow{E}=200cos53\hat{i}+200sin53\hat{j}$
$=(120\hat{i}+160\hat{j})N{C}^{-1}$