Question

Find the net electric field due to a finite wire of linear charge density $+\lambda \text{C/m}$ at a point $\text{P}$ located at a perpendicular distance $d$ as shown in the figure.

$[K=\frac{1}{4\pi {\epsilon }_0}]$

• A. $\sqrt{2}\frac{K\lambda }{d}$
• B. $2\sqrt{2}\frac{K\lambda }{d}$
• C. $\frac{K\lambda }{\sqrt{2}d}$
• D. $\frac{1}{2\sqrt{2}}\frac{K\lambda }{d}$

Answer 1

The correct option is A $\sqrt{2}\frac{K\lambda }{d}$


We know that electric field intensity $\left(E_{perp}\right)$ perpendicular to the uniformly charged wire at distance $d$ is,

$E_{perp}=\frac{K\lambda }{d}[\sin {\theta }_1+\sin {\theta }_2]$

According to problem,

${\theta }_1={30}^{\circ }$ and ${\theta }_2={60}^{\circ }$

Substituting the values in the above equation, we get

$E_{perp}=\frac{K\lambda }{d}[\sin {30}^{\circ }+\sin {60}^{\circ }]$

$\Rightarrow E_{perp}=\frac{K\lambda }{d}[\frac{1}{2}+\frac{\sqrt{3}}{2}]$

$\Rightarrow E_{perp}=\frac{K\lambda }{d}\left[\frac{\sqrt{3}+1}{2}\right]$

Electric field intensity $\left(E_{para}\right)$ parallel to the uniformly charged wire at distance $d$ is,

$E_{para}=\frac{K\lambda }{d}[\cos {\theta }_1-\cos {\theta }_2]$

Substituting the values of ${\theta }_1$ and ${\theta }_2$, we get,

$E_{para}=\frac{K\lambda }{d}[\cos {30}^{\circ }-\cos {60}^{\circ }]$

$\Rightarrow E_{para}=\frac{K\lambda }{d}[-\frac{1}{2}+\frac{\sqrt{3}}{2}]$

$\Rightarrow E_{para}=\frac{K\lambda }{d}\left[\frac{-1+\sqrt{3}}{2}\right]$

Now, net electric field intensity at point $\text{P}$,

$E_{\text{net}}=\sqrt{(E_{para}{)}^2+(E_{perp}{)}^2}$

$\Rightarrow E_{\text{net}}=\sqrt{{\left(\frac{K\lambda }{d}\right)}^2[{\left(\frac{1+\sqrt{3}}{2}\right)}^2+{\left(\frac{-1+\sqrt{3}}{2}\right)}^2]}$

$\Rightarrow E_{\text{net}}=\frac{K\lambda }{2d}\sqrt{(1+\sqrt{3}{)}^2+(-1+\sqrt{3}{)}^2}$

$\Rightarrow E_{\text{net}}=\frac{K\lambda }{2d}\sqrt{4+4}$

$\therefore E_{\text{net}}=\sqrt{2}\frac{K\lambda }{d}$

Hence, option (a) is the correct answer.