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) \[\Rightarrow \] \[{{\varepsilon }_{0}}E\times 2\pi rh=\lambda h\] \[\Rightarrow \] \[E=\frac{1}{2\pi {{\varepsilon }_{0}}}.\frac{\lambda }{r}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\]You need to login to perform this action.
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