$\frac1{4πε_0} \frac{λ}{r}$
$\frac1{4πε_0} \frac{λ}{r^2}$
$ \frac{λ}{2ε_0}$
$ \frac{λ}{2\piε_0 r}$
The correct answer is (D): $ \frac{λ}{2\piε_0 r}$
Consider an infinity long straight conductor of linear charge density
Consider a cylindrical Gaussian surface of radius r and length l.
charge enclosed by the surface
\(q=λl\)
By Gauss law
$\phi \vec{E}.ds = \frac{q}{ε_0} = \frac{λl}{ε_0} $
$ \vec{E}\phi ds= \frac{λl}{ε_0} $
$ \vec{E} \times 2\pi rl = \frac{λl}{ε_0} $
$ E = \frac{λ}{2\pi rε_0}$
A cubical volume is bounded by the surfaces $x=0, x= a , y=0, y= a , z=0, z= a$ The electric field in the region is given by $\vec{E}=E_0 x \hat{ t }$ Where $E_0=4 \times 10^4 NC ^{-1} m ^{-1}$ If $a=2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$ The value of $Q$ is ___ Take \(E_{0}=9\times 10^{-2}C^{2}/Nm^{2}\)