Question

For a finite line charge, find the electric field at the non-axial point $\text{P}$, perpendicular to axis of line charge.

• A. $44.6\text{N/C}$
• B. $57.6\text{N/C}$
• C. $60.6\text{N/C}$
• D. $63.6\text{N/C}$

Answer 1

The correct option is B $57.6\text{N/C}$

Given:
$r=4\text{m};\lambda =20\text{nC/m}=20\times {10}^{-9}\text{C/m}$

For Pythagoras theorem, in $\Delta \text{APC}$ and $\Delta \text{BPC}$,

$\text{AP}=2\sqrt{13}\text{m}$ and $\text{BP}=2\sqrt{5}\text{m}$

So, $\sin \alpha =\frac{3}{\sqrt{13}}$ and $\sin \beta =\frac{1}{\sqrt{5}}$

We know that, the electric field at the non-axial point $\text{P}$ perpendicular to the line charge axis

$E_{\perp }=\frac{kQ}{Lr}[\sin \alpha +\sin \beta ]$

$\Rightarrow E_{\perp }=\frac{k\lambda }{r}[\sin \alpha +\sin \beta ](\because \lambda =Q/L)$

$\Rightarrow E_{\perp }=\frac{9\times {10}^9\times 20\times {10}^{-9}}{4}[\frac{3}{\sqrt{13}}+\frac{1}{\sqrt{5}}]$

$\Rightarrow E_{\perp }=57.6\text{N/C}$

Hence, option $\left(b\right)$ is the correct choice.