Question

At the moment `t = 0` on electron leaves one plate of a parallel-plate capacitor with a neglible velocity. An accelerting volatage, varrying as `V = at`, where `a = 100 V//s` is applied between the plates is `l = 5.0 cm`. What is the velocity of the of the electron at the moment it reaches the opposite plate?

Answer 1

Let us electron leave the negative plate of the capacitor at time `t = 0`
As, `E_(x) = (d varphi)/(d x), E = (varphi)/(l) = (a t)/(l)`,
and, therefore, the acceleration of the electron,
`w = (eE)/(m) = (eat)/(ml)` or, `(dv)/(dt) = (eat)/(ml)`
or, `int_(0)^(v) dv = (ea)/(ml) int_(0)^(t) tdt`, or, `v = (1)/(2) (ea)/(ml) t^(2)` .....(1)
But, from `s = int v dt`,
`l = (1)/(2) (ea)/(ml) int_(0)^(t) = (eat^(3))/(6 ml)` or, `t = ((6ml^(2))/(ea))^(1/3)`
Putting the value of `t` in (1),
`v = (1)/(2) (ea)/(ml) ((6ml^(2))/(ea))^(2/3) = ((9)/(2) (al e)/(m))^(1/3) = 16 km//s`