Accelerating Protons in a Particle Accelerator

[pokeefer]

Homework Statement

Protons are being accelerated through a potential difference of 5,000 V in the gun of a particle accelerator. They must pass between two parallel deflecting plates that are 0.06 m long and 0.02 m apart with a potential difference between them of 1,500 V

Homework Equations

Ep = kQ1Q2 / d^2
E= V/d
F = QE
V = Change in Ep / Q
I = Q/t
V = kQ / d

The Attempt at a Solution

First part asks me:

What is the kinetic energy of protons leaving the gun?

What do i do?

 

[Delphi51]

In all these acceleration-through-potential-difference problems, there are just two approaches. The first has all the details:
potential dif causes E field which causes F on electrons which causes them to accelerate which increases their velocity. You can write a basic formula for each "causes" and combine them in a chain to find any of those quantities given any other.

The second approach is a big shortcut that often finds what you want. Use the definition of potential difference: it is the energy given (or taken) from each charge, V = Energy/Q. Quite often you want the speed of the charge so you replace Energy with the KE formula.

 

[pokeefer]

So would I do something like this?

V = Energy Potential / Q

V = Energy Kinetic / Q

5,000 v = Energy Kinetic / (1.6 x 10^-19)

Then to find velocity you do:

Energy Kinetic = 1/2 mv^2

And in the mass you put the mass of the proton (1.67 x 10^-27) ?

Just want to make sure.

 

[Delphi51]

It all sounds good!

 

[rwisz]

To calculate the kinetic energy of the protons, we can use the equation K = mv^2/2, where m is the mass of the proton and v is its velocity. In order to calculate the velocity, we need to use the equations for electric potential energy (Ep) and electric field (E).

First, we can calculate the electric potential energy of the protons as they pass through the gun using the equation Ep = kQ1Q2/d^2, where k is the Coulomb's constant, Q1 is the charge of the proton (1.6 x 10^-19 C), Q2 is the charge of the gun (also 1.6 x 10^-19 C), and d is the distance between them (0.06 m). This gives us an Ep of 2.67 x 10^-14 J.

Next, we can calculate the electric field between the two parallel plates using the equation E = V/d, where V is the potential difference between the plates (1,500 V) and d is the distance between them (0.02 m). This gives us an E of 75,000 V/m.

Now, we can use the equation for electric field (F = QE) to calculate the force acting on the protons as they pass through the plates. This gives us a force of 1.2 x 10^-15 N.

Using the equation for work (W = Fd), we can calculate the work done on the protons as they pass through the plates. This gives us a work of 2.4 x 10^-17 J.

Finally, we can use the equation for kinetic energy (K = W = mv^2/2) to solve for the velocity of the protons leaving the gun. Plugging in the mass of a proton (1.67 x 10^-27 kg) and the work we calculated, we get a velocity of 2.09 x 10^7 m/s.

Therefore, the kinetic energy of the protons leaving the gun is 1.74 x 10^-10 J.