How far has the proton been deflected?

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In summary, the proton's horizontal velocity is constant, so to find its vertical velocity and therefore how far it is deflected, we need to find the acceleration using a= eE/m. The surface charge densities on the plates of the capacitor are given, and using Gauss's law, we can find the electric field using E=sigma/epsilon_0. The units must match for this to work. Sigma is the electric charge surface density on the plates, and the area can be found from the given information. A PDF file can be downloaded to learn more about using TeX editing for equations.
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A proton traveling 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.
 
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  • #2
I believ the electric field is
[tex] E=\frac{\sigma}{\epsilon_{0}} [/tex]

Gauss theorem could prove that...

Daniel.
 
  • #3
dextercioby said:
I believ the electric field is
[tex] E=\frac{\sigma}{\epsilon_{0}} [/tex]

Gauss theorem could prove that...

Daniel.
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 [tex] ... [/tex] but those images don't show up, only the texts were italicized and bold.
 
  • #4
1.Sorry,the units have to match...
[tex] \nabla\cdot (\epsilon_{0}\vec{E})=\rho [/tex]

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
dextercioby said:
1.Sorry,the units have to match...
[tex] \nabla\cdot (\epsilon_{0}\vec{E})=\rho [/tex]

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.

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
(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.
 

Related to How far has the proton been deflected?

1. How is the proton's deflection measured?

The proton's deflection is measured using a device called a particle accelerator, which uses strong electric and magnetic fields to deflect the proton's path. The amount of deflection is then calculated using the known properties of the electric and magnetic fields.

2. What is the significance of the proton's deflection?

The proton's deflection is significant because it allows us to study the properties of subatomic particles and the fundamental forces that govern their behavior. By observing how protons are deflected, we can gain a better understanding of the structure of matter and the laws of physics.

3. How far can a proton be deflected?

The distance a proton can be deflected depends on the strength of the electric and magnetic fields used in the particle accelerator. In most cases, protons can be deflected up to a few centimeters, but in high-energy accelerators, they can be deflected for much longer distances.

4. What factors affect the proton's deflection?

The main factors that affect the proton's deflection are the strength of the electric and magnetic fields, the speed of the proton, and the mass of the proton. Additionally, the shape and design of the particle accelerator can also have an impact on the proton's deflection.

5. How does the proton's deflection relate to other subatomic particles?

All subatomic particles, including protons, can be deflected using particle accelerators. However, the amount of deflection may vary depending on the particle's charge, mass, and speed. For example, electrons are much lighter than protons and therefore are deflected more easily, while neutrons have no charge and are not deflected by electric fields.

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