Question

Consider an electron beam of uniform circular cross-section with a diameter of 1 mm. The electrons are accelerated from the cathode to the anode, which are separated by 1 m. The electrons are generated at the cathode and therefore have negligible velocity at this position. The electrons accelerate at a rate of 108 ?/?2 between the cathode and anode. The current density at the anode is 104 ?/?2 .

Find the current density as a function of position between the cathode and anode.

Answer 1

NOTE - We are considering only magnitude of current density here, not the direction. Its direction is opposite to the direction of motion of electrons, i.e., negative x direction.

According to the problem ==>

Distance between electrodes, d = 1 (m)

Let the position of cathode be x= 0, therefore, position of anode be x = d = 1. Any position in between can simply be called as x (in m).

Therefore, velocity at cathode, v_c = v(0) = 0

Acceleration of electrons = a = 108 ?/?²

Let's find velocity at anode, v_a and velocity in between the electrodes at position x, v(x) ==>

Therefore, Velocity at anode, v_a = v(d) ==>

Therefore, v(x) ==>

Let number of electrons per unit area = n (/m²)

Also, elementary charge on the electron = q (C)

Therefore, current density at anode, J_a = J(1) = n*q*v_a = 104 (?/?² )

Therefore, current density at x, J_x = J(x) = n*q*v_x = n*q*v_a * x^0.5 = 104 * x^0.5 (?/?² )