Question

If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.

Answer 1

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

I = q*N

Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron,
`q = 1.602 * 10^(-19)` Coulombs, then the current is

`I = (1.602*10^(-19))(10^(15))`

`I = 0.1602 mA`

And the power in the Beam is

`P = VI`

`P = (350*10^3)(0.1602)`

`P = 0.05607 Watts`