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

$P = VI$

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

$P = 0.05607 Watts$