An Empire starship fires at Skywalker’s ship

An Empire starship fires a beam of electrons at Skywalker’s ship. Each electron has an initial kinetic energy of 1.6×10−17 J. The Skywalker’s ship has been fitted out with the electrostatic shield technology for defense against electron beam weapons. The electric field of the shield points directly inward. If the electrons stop after travelling 83 m, what is the size of the electric field generated by the ship’s shield? Please ignore relativistic effects.

Particle enters a magnetic field. Question 3.

An electron is accelerated from rest through a potential difference of 200 V and then enters a magnetic field of strength 0.02 T acting at right angles to its path. Calculate the radius of the resulting electron orbit.

Solution.

Based on the Law of Conservation of Energy

Ep – Electric potential energy ; Ek – Mechanical Kinetic energy

Ep = Ek

Ep = q*U ; U=200 v

Ek = m * v^2 / 2

q*U = m * v^2 / 2

Now we could find a velocity. v = SQRT(2*q*U/m)

Lorentz force. It is a magnetic force on a point charge.

F = q * v* B * sin (A)

F – centripetal force equal :

F= ma = m * v^2/r = q * v* B * sin (A), sin (90) = 1

Now, we could find another equation for the velocity

v = q*B*r/m

Finally, we have everything in order to calculate radius.

We also could use this equations and get another equation.

v=SQRT(2*q*U/m)
v = q*B*r/m

So, r= SQRT(2*U*m/(q*B^2) ) = SQRT (2*200*9.1*10^-31/(1.6*10^-19*0.02^2)) = 0.002 m