# Field of charged hollow cylinder (Griffiths Physics)

There is a surface charge density $rho_{s0}$on the surface of a cylinder $rho=a$, $0 < z < L$, $0 < phi < 2pi$. What is the electric field $overrightarrow{E}$ at the point $(0,0,h)$

First compute the electric field of a charged disk (see above) at point P at distance z on the axis

$d E_P=(k*d q)/d^2 =k d q/sqrt{a^2+z^2}$     and
$d E_{z P}=d E_P*cos⁡ θ=k*d q/(a^2+z^2 )*z/sqrt{a^2+z^2}$

For the same z the entire field of the ring at point P is just

$E_P=k z/(a^2+z^2 )^{3/2} ∫dq=k z q/(a^2+z^2 )^{3/2}$

We cut this ring from the cylinder:

$q=ρ*d S=ρ*2πa*dz$

We have finally the ring cut from the disk as in the second figure. Field of the ring is

$E_P=(k q (h-z))/(a^2+(h-z)^2 )^{3/2}$

Finally we integrate the above filed along all the cylinder length

$E_p=2πaρ*k∫_0^L(h-z)dz/[a^2+(h-z)^2 ]^{3/2} =-aρ/(2ϵ_0 ) ∫_h^{h-L} t d t/[a^2+t^2 ]^{3/2}=$ $=aρ/(2ϵ_0 ) 1/sqrt{a^2+t^2} (from (h) to (h-L))$

$E_P=aρ/2ϵ_0 [1/sqrt{(a^2+(h-L)^2 )}-1/sqrt{(a^2+h^2 )}]$