Question

For a finite line charge, the net electric field at the non-axial point $\text{P}$ is directed toward the line $\text{PQ}$ where $\text{PQ}$ makes an angle $\theta$ with the axis of line charge. Find the value of $\theta$.

• A. ${\tan }^{-1}\left(10\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(14\right)$
• D. ${\tan }^{-1}\left(26\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(14\right)$


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

$\text{AP}=5\text{m}$ and $\text{BP}=3\sqrt{2}\text{m}$

So, $\sin \alpha =\frac{4}{5},\cos \alpha =\frac{3}{5},\sin \beta =\frac{1}{\sqrt{2}}$ and $\cos \beta =\frac{1}{\sqrt{2}}$

We know that, the electric field at the non-axial point $\text{P}$,

$E_{\perp }=\frac{kQ}{Lr}[\sin \alpha +\sin \beta ]......\left(1\right)$

$E_{||}=\frac{kQ}{Lr}[\cos \beta -\cos \alpha ].......\left(2\right)$

Now, $\tan \theta =\frac{E_{\perp }}{E_{||}}$

So, from eqs. (1) and (2)

$\Rightarrow \tan \theta =\frac{\sin \alpha +\sin \beta }{\cos \beta -\cos \alpha }$

$\Rightarrow \tan \theta =\frac{\frac{4}{5}+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-\frac{3}{5}}$

$\Rightarrow \tan \theta =14$

$\Rightarrow \theta ={\tan }^{-1}\left(14\right)$

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