Answer:
Explanation:
Given
Length of shell [tex]L=220\ m[/tex]
radius of cylindrical shell [tex]r=4\ cm[/tex]
surface charge density [tex]\sigma =14\ nC/m^2[/tex]
Total charge on the shell [tex]Q=surface\ area\times surface\ charge\ density[/tex]
surface area [tex]A=2\pi r\cdot L=2\pi \cdot 0.04\cdot 220=55.299\ m^2[/tex]
[tex]Q=\sigma \cdor A[/tex]
[tex]Q=14\times 10^{-9}\times 55.299[/tex]
[tex]Q=7.741\times 10^{-7}\ C[/tex]
