Proton in a plate capacitor: how far does it go?

[pickleparty]

1. Homework Statement

A proton traveling at a speed of 7.22 *106m/s enter the gap between the plates of a 7.86 cm wide parallel plate capacitor. The surface charge densities on the plate are ± 4.98 *10-6 C/m2. How far (in m) has the proton deflected sideways when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

2. Homework Equations

S= (1/2)*(E*q)/m*(d/v)^2 ?

3. The Attempt at a Solution

I just don't know what goes where or if I even have the right formula. I am truthfully lost. A classmate told me to use the above formula but didnt explain it. I got the concept he explained but lost the math. so what formula and how does it fit together?

 

[ideasrule]

Can you find the electric field inside the capacitor? If so, can you find the acceleration of the electron? With the acceleration and the time the electron takes to transverse the capacitor, can you find distance deflected?

 

[ehild]

The proton travels along the length of the capacitor (x direction)with the given speed v_x and is deflected by the electric field in the y direction. Like in case of projectile motion we can consider separately the motions along x and y: in the x direction it is uniform x=v_xt , in the y direction the proton moves with uniform acceleration: y=1/2 *a *t^2.
You need t: but you know that x= 7.86*10-2 m and v_x=7.22 *10^6 m/s.
You need the acceleration a. It is force/mass. How much is the force on a proton in an electric field E? You are given the surface charge of the capacitor. How is it related to the electric field strength? To get the force, you need the charge of proton. To get the acceleration, you need the mass of the proton. Both data are included in your textbook, or you can find them on the Net.

ehild

 

[pickleparty]

I'm still missing something. can you dumb it down more. do i use the formula my friend told me to use or do i ues another
mass proton: 1.6*10^-27
charge of proton: 1.6*10^-19 C
S= (0.5 E?*q?)/(1.6*10^-27((4.98 *10-6 C/m2)/(7.22 *106m/s))^2)

 

[ehild]

Do not use a formula without understanding it.

Try to answer my questions. I help a bit. The electric field E between the capacitor plates is equal to the surface charge density on the plates divided by the vacuum permittivity.



ehild