question_answer

The magnitude of electric field at distance r from an infinitely thin rod having a linear charge density \[\lambda \]is (use Gausss law)

A) \[E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\]                               

B) \[E=\frac{2\lambda }{2\pi {{\varepsilon }_{0}}r}\]

C) \[E=\frac{2\lambda }{4\pi {{\varepsilon }_{0}}r}\]                            

D) \[E=\frac{4\lambda }{\pi {{\varepsilon }_{0}}r}\]

Correct Answer: A

Solution :

                Gauss theorem states that the total electric flux\[(\phi )\]through any closed surface is equal to\[\frac{1}{{{\varepsilon }_{0}}}\]times the net charge q enclosed by the surface. \[{{\phi }_{E}}=\int{\overrightarrow{E}}.\overrightarrow{dA}=\frac{Q}{{{\varepsilon }_{0}}}\] 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\[=\lambda h\] \[\therefore \] From Gauss theorem, Charge\[={{\varepsilon }_{0}}\times \](Electric flux) \[\Rightarrow \]               \[{{\varepsilon }_{0}}E\times 2\pi rh=\lambda h\] \[\Rightarrow \]               \[E=\frac{1}{2\pi {{\varepsilon }_{0}}}.\frac{\lambda }{r}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\]