How Fast Must a Proton Travel to Cross an Electric Field?

[Foxhound101]

Homework Statement

Two 4.0 cm diameter disks face each other, 2.0 mm apart. They are charged to +-12 Nc.

Part A
What is the electric field strength between the disks?
E = 1.1 * 10^6 N/C

Part B
A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

Homework Equations

E=(Q/(epsilon zero * A)
v=x/t

The Attempt at a Solution

I was able to figure out part A. I just need help with part B.
V = ?
x = 2mm
t = ?

Doesn't seem like this part should be difficult, so I must be missing something simple...

Thanks in advance for any help.

 

[Doc Al]

Two ways to attack this:

(1) Using kinematics. Hint: What's the proton's acceleration?

(2) Using energy. Hint: What's the potential difference between the plates?

 

[Foxhound101]

So...this is what I have so far.

F = qE

F = (12*10^-9)(1.1*10^6)
F = (.0132N)

F = ma

mass of proton = 1.67*10^-27

.0132 = (1.67*10^-27) (a)
7.9*10^24

 

[Doc Al]

So...this is what I have so far.

F = qE

F = (12*10^-9)(1.1*10^6)
F = (.0132N)

Since you need the force on the proton, use the charge of a proton. (Not the total charge on the plate!)

 

[Foxhound101]

F=(1.6*10^-19)(1.1*10^6)
F = 1.76*10^-13

F=ma

1.76*10^-13 = (1.67*10^-27)(a)
1.05*10^14 = a

 

[Doc Al]

Once you've found the acceleration, it's time for kinematics. You'll need a kinematic equation relating speed and distance.

 

[Foxhound101]

v = x/t
f = m/a

Those aren't it...hm...

Kinetic energy = .5(mass)(velocity)^2

so...(if I remember correctly) total energy = Kinetic energy + potential energy

potential energy = (mgh)

Sadly, if this is the correct approach I do not remember what variables are on the total energy side.

*edit*
O yeah...I forgot I was looking for an equation relating speed and distance. I am having trouble finding one.

 

[Foxhound101]

Hm...perhaps this equation
v^2 = vo^2 + 2a(X - Xo)

*edit*
Yup...that would be the correct equation.

Thanks for the help Doc Al.