# How far has the proton been deflected?

1. Jan 23, 2005

### Heart

A proton travelling at v= 1.0*10^5 m/s enters the gap btwn the plates of a 2.0-cm-wide parallel-plate capacitor. The surface charge densities on the plates are +-1.0*10^6 C/m2. How far has the proton been deflected sideways when it reaches the far edge of the capacitor?

I figured that the horizontal velocity would be constant so I need to find a vertical v to find how far the proton has been deflected. After that I can find the answer.

So, I need to find the acceleration to find the final vertical v by using a= eE/m.

The problem is I'm not sure how to find E in N/C w/ the given information. They give 1.0*10^6 C/m2 but I'm not sure how to convert that into E.

Any hints/tips would be greatly appreciated.

2. Jan 23, 2005

### dextercioby

I believ the electric field is
$$E=\frac{\sigma}{\epsilon_{0}}$$

Gauss theorem could prove that...

Daniel.

3. Jan 23, 2005

### Heart

Thanks for the reply, but if you don't mind, can you clarify a bit?

I know what epsilon_0 is but not sigma. My class hasn't reached that part yet. If sigma is given 1.0*10^-6 C/m^2, it still doesn't make sense on how E = sigma/epsilon_0; I tried but the units don't work out right.

BTW, if you don't mind, can you teach me how to get those epsilon, sigma, et cetera images? I tried copying your $$...$$ but those images don't show up, only the texts were italicized and bold.

4. Jan 23, 2005

### dextercioby

1.Sorry,the units have to match...
$$\nabla\cdot (\epsilon_{0}\vec{E})=\rho$$

is Gauss law for vacuum...Your formula results immediately.

2.sigma is te electric charge surface density on the plates of the capacitator.U could have figured that after the units (C/m^{2}).

3.Yes,u mean TeX editing.Dld the file which teaches u how to write the code lines.Click on any of the formulas and then another click in the small window that pops up.It's a PDF file.

Daniel.

5. Jan 23, 2005

### Heart

Daniel, I'm lost here. I still don't get what can I use sigma for exactly? Sigma is the electric charge/area. Only the length is given, how can I find the electric field from sigma then when I have no clue what the area is?

Also, in the above reply, I see that you use rho in the equation, but rho is not given at all. I'm really confused; if you don't mind, can you clarify a bit?

Also in the above reply, you used an upside down delta, what is that thing?

6. Jan 23, 2005

### krab

(in words) the electric field is equal to the surface charge density divided by the permittivity of free space. You have both numbers, so just divide them.