Question

A system designed to accelerate and deflect protons can be composed of two sets of charged parallel plates, the first with the plates aligned parallel to the y-axis and the second parallel to the x-axis.

a. If the first section has a potential difference of 50 kV across its 6 mm gap what velocity (vx) will the proton achieve when it crosses the gap between the plates. (Hint: use kinetic energy)

b. The proton then passes into center of the gap (also 6 mm wide) of the second set of plates where it experiences a total deflection of 1.2 mm in the +y-direction. Assuming the E field is uniform between the plates what force does the proton experience while between these plates (Note: initially vy=0)

c. What is the direction and magnitude of the field between these plates

d. What is the surface charge density on these plates

Answer 1

a) The potential difference between the plates =50 KV

Gap between the plates = 6 mm

Here, the gap between the two plates is very small compared to the plates dimensions.

The force acting on the proton placed in the field is F = qE ( Since E = V/d)

=> ma = qE

=> a = (q/m)*E

Now , the velocity of the particle is given as V² = 2ax

(x is the distance between t he two plates)

=> V =

so V = 9.58*(10¹²) m/s

b)

c) The field direction must be opposite to the proton direction. So, the field direction is towards the +x-axis.

d) The surface charge density on the plates is given by the equation

Then surface charge density 1.475 * 10^-4 kg/m²