question_answer

The electric field \[E={{E}_{0}}y\hat{j}\] acts in teh space in which a cylinder of radius r and length \[l\] is placed with its axis parallel to y-axis. The charge inside the volume of cylinder is :

A) \[{{E}_{0}}{{\varepsilon }_{0}}\frac{{{l}^{2}}}{2}\]

B) \[{{E}_{0}}{{\varepsilon }_{0}}\pi {{r}^{2}}{{l}^{2}}\]

C) \[{{E}_{0}}{{\varepsilon }_{0}}\pi {{r}^{2}}l\]   

D) \[2{{E}_{0}}{{\varepsilon }_{0}}\pi {{r}^{2}}l\]

Correct Answer: C

Solution :

Flux through face (1) (entering)\[=-{{E}_{0}}{{y}_{0}}\pi {{r}^{2}}\].    Flux through face (2) \[0+0=C\Rightarrow \,C=0\]                 Net Flux\[={{E}_{0}}\pi {{r}^{2}}l=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}\]                 \[{{q}_{in}}={{E}_{0}}{{\varepsilon }_{0}}\pi {{r}^{2}}l\]