question_answer

An infinitely long thin straight wire has uniform linear charge density of\[\frac{1}{3}C{{m}^{-1}}.\]. Then, the magnitude of the electric intensity at a point 18 cm away is \[(given\,{{\varepsilon }_{0}}=8.8\,\times \,{{10}^{-12}}\,{{C}^{2}}\,N{{m}^{-2}})\]

A)  \[0.33\times {{10}^{11}}N{{C}^{-1}}\]

B)  \[3\times {{10}^{11}}N{{C}^{-1}}\]

C)  \[0.66\times {{10}^{11}}N{{C}^{-1~~}}\]

D)  \[1.32\times {{10}^{11}}N{{C}^{-1}}\]  

Correct Answer: A

Solution :

Charge density of long wire            \[\lambda =\frac{1}{3}C-m\] and \[r=18\times {{10}^{-2}}m\]   From Gauss theorem              \[\oint{\overrightarrow{E}.d\overrightarrow{S}=\frac{q}{\frac{{{e}_{0}}}{{}}}}\] \[E\oint{dS=\frac{q}{{{\varepsilon }_{0}}}}\] or \[E\times 2\pi rl=\frac{q}{{{\varepsilon }_{0}}}\] or \[E=\frac{q}{2\pi {{\varepsilon }_{0}}rl}=\frac{q/l}{2\pi {{\varepsilon }_{0}}r}\] \[=\frac{\lambda \times 2}{2\pi {{\varepsilon }_{0}}r\times 2}=\frac{\lambda \times 2}{4\pi {{\varepsilon }_{0}}r}\] \[=9\times {{10}^{9}}\times \frac{1}{3}\times 2\times \frac{1}{18\times {{10}^{-2}}}\] \[=\frac{1}{3}\times {{10}^{11}}=0.33\times {{10}^{11}}\] \[0.33\times {{10}^{11}}N{{C}^{-1}}\]