Q. The magnitude of electric field at distance $r$ from an infinitely thin rod having a linear charge density $\lambda$ is(use Gauss’s law)
A) $E = \frac{\lambda}{2\pi\varepsilon_{0}r}$
B) $E = \frac{2\lambda}{\pi\varepsilon_{0}r}$
C) $E = \frac{\lambda}{4\pi\varepsilon_{0}r}$
D) $E = \frac{4\lambda}{\pi\varepsilon_{0}r}$
Explanation
(a) According to Gauss theorem for closed surface
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Taking cylindrical Gaussian surface of radius $r$, height $h$ curved surface=$2\pi rh$.
Electric flux through it =$E \times 2\pi rh$
Charge enclosed=$\lambda h$
∴ From Gauss theorem,
$Charge = \varepsilon_{0} \times (Electric\ flux)$
$\Longrightarrow \varepsilon_{0}E \times 2\pi rh = \lambda h$
$\Longrightarrow E = \frac{1}{2\pi\varepsilon_{0}}.\frac{\lambda}{r} = \frac{\lambda}{2\pi\varepsilon_{0}r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.