A homongeous proton beam acceletated by a potentiail difference `V = 600 kV` has a round cross-section of radius `r = 5.0 mm`. Find the electric fiel dstrength on the surface of the beam and the potentia difference between the surface of the beam and the potential difference between the surface and the axis of the beam if the beam current is equal to `I = 50 mA`.

Explanation

From Gauss theroem field strength at a surface of a cylindirical shape equals, `(lambda)/(2pi epsilon_(0) r)`, where `lambda` is the inear charge density.
Now, `eV = (1)/(2) m_(e) V^(2)` or, `v = sqrt((2eV)/(m_(e)))` ....(1)
Also, `dq = lambda dx` so, `(dq)/(dt) = lambda (dx)/(dt)`
or, `I = lambda v` or, `lambda = (I)/(v) = (1)/(sqrt((2eV)/(m_(e))))`, using (1)
Hence, `E = (I)/(2pi epsilon_(0) r) sqrt((m_(e))/(2 e V)) = 32 V//m`
(b) For the point, inside the solid charged cyclinder, applying Gauss's theorem,
`2 pi r h E = pi r^(2) h (q)/(epsilon_(0) pi R^(2) l)`
or, `E = (q//l)/(2pi epsilon_(0) R^(2))`
So, from `E = - (d varphi)/(dr)`,
`int_(varphi_(1))^(varphi_(2)) -d varphi = int_(0)^(R) (lambda)/(2pi epsilon_(0) R^(2)) r dr`
or, `varphi_(1) - varphi_(2) = (lambda)/(2pi epsilon_(0) R^(2)) [(r^(2))/(2)]_(0)^(R) = (lambda)/(4pi epsilon_(0))`
Hence, `varphi_(1) - varphi_(2) = (VI)/(4pi epsilon_(0)) sqrt((m_(e))/(2e V)) = 0.80 V`