Question

Hi everyone! What is the electric field vertically above a standing hollow uniformly charged cylinder?
Pls help I really don't know!!

Answer 1

\[dq=2\pi R \sigma dx\]
\[dE_{v}=\frac{ dq }{ 4\pi \epsilon_0 r^2 }\cos \theta=\frac{ R \sigma (x+h)dx }{2 \epsilon_0 r^3 }=\frac{ R \sigma (x+h)dx }{2 \epsilon_0 ((x+h)^2+R^2)^\frac{ 3 }{ 2 } }\]
\[E_v=\frac{ R \sigma }{2 \epsilon_0 }\int\limits_{0}^{l}\frac{(x+h)dx }{((x+h)^2+R^2)^\frac{ 3 }{2 }}\] where l is the lenght of the cylinder
\[E_v= \frac{ R \sigma }{ 2 \epsilon_0 }(\frac{ 1 }{ \sqrt{h^2+R^2} }-\frac{ 1 }{ \sqrt{(h+l)^2+R^2} })\]