Question about Particle with Velocity entering an Electric Field

[iiguitar]

Homework Statement

A Proton with a velocity of 1.8x10^5 m/s and mass of 1.673x10^27 kg approaches an electric field created by two plates 30 cm apart. The electric field extends length wise 15 cm across the page with a potential difference of 50,000V. How far will the proton be deflected? (Distance of deflection from original position perpendicular to the plates 30cms apart)

Homework Equations

The Attempt at a Solution

Not quite sure but I think that if the question were to give a magnetic field, we would use
F=qvBsin(*)=(mv^2)/r but its not :(

So I figured that the the proton will be deflected towards the negative plate with Kinetic energy of 1/2mv^2. This KE would be equal to the Work done that is W=Fd

But the force of deflection is equal to Eq. I can work out E from the voltage(50,000V) divided by the distance of the plates (30cm =0.3m) times it by 1.6x10^-19 C ( charge of proton)

So far, with that working out, I have this result:

KE = 1/2mv^s = 0.5 x 1.673x10^-27 x (1.8x10^5)^2 = 2.71026x10^-17

KE= 2.7x10^-17 = W

W= Fd

F= Eq = 50000/0.3 x 1.6x10^-19 = 2.667x10^-14 N

so 2.7x10^-17 = 2.667x10^-14 x d

Therefore distance of deflection =
2.7x10^-17/2.667x10^-14
=1.0163x10^-3 m
equal approximately 0.1 cm?

I'm not sure if what I've done is correct. Can some please verify my results and method of approaching this question please? If anything is unclear please ask so I can clear it up.

Thanks :)

 

[ideasrule]

Before getting to your calculations, I have to say that the question might be ambiguous. I think it means "when the proton is about to exit the space between the plates, how much closer is it to the negative plate than when it entered?"

KE = 1/2mv^s = 0.5 x 1.673x10^-27 x (1.8x10^5)^2 = 2.71026x10^-17

KE= 2.7x10^-17 = W

W= Fd

F= Eq = 50000/0.3 x 1.6x10^-19 = 2.667x10^-14 N

so 2.7x10^-17 = 2.667x10^-14 x d

No, because the electric field does not give the proton all its kinetic energy. The proton had plenty of kinetic energy to begin with, and you'll probably find that the work the field does is very small compared to this initial energy.

You can (and should) do this problem using only forces. Specifically, can you write out the equations of motion for the x and y axis? If so, when x=15 cm (assuming the x-axis is perpendicular to the electric field), what's y? What would y be if there was no field?

 

[AlexChandler]

I would first find the time it takes the proton to travel the total distance through the plates (using the initial speed given, it should just be

$$t = \frac{distance}{speed_x}$$

Call the line along which the proton initially moves, the y axis. And the x-axis is perpendicular to this direction.

Then you know the acceleration of the proton by

$$qE = ma_x$$

You find that

$$a_x = qE/m = const$$$$v_x = a_x t$$

and

$$x= \frac{1}{2} a_x t^2$$

Using the t you solved for at the start you should find how far the proton travels in this direction.

Where you have gone wrong is that the work done is equal to the $$change$$ in kinetic energy, not the $$initial$$ kinetic energy.

I think that it should be easier to solve it this way instead of trying to use energy.