How do I use Gauss's Law to calculate proton trajectories in an electric field?

In summary, The student is having difficulty with a physics problem and asks for help. He states that he is new to the subject and does not understand the equations. He is trying to solve for two projections angles that will result in a hit with a proton. He is using a projectile range equation to figure out his acceleration. Once he knows g, he can solve for θ.
  • #1
Gaara09
9
0

Homework Statement


Hey , I am a new member Here . I am having some diffuculties , i did one problem and still 1 problem i can't do it coz it seems very hard as i am new to gauss's Law

I am doing college physics II , protons are projected with initial speed of v0 = 9550m/s into a region in which a uniform electric field , E= 720N/C is present . the protons are to hit a target that lies a horizental distance of 1.27mm from the point at which the protons are lauched.
Find the two projections angles that will result in the hit .

b- the total duration of flight for each of thse two trajectoires.
1zqgeo1.jpg



The Attempt at a Solution


I tried to do these E= K.(q)/r^2 , E.A.cos (x). well i don't have any clue...

I hope you can help me

Thanks
 
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  • #2
Gaara09 said:

Homework Statement


Hey , I am a new member Here . I am having some diffuculties , i did one problem and still 1 problem i can't do it coz it seems very hard as i am new to gauss's Law

I am doing college physics II , protons are projected with initial speed of v0 = 9550m/s into a region in which a uniform electric field , E= 720N/C is present . the protons are to hit a target that lies a horizental distance of 1.27mm from the point at which the protons are lauched.
Find the two projections angles that will result in the hit .

b- the total duration of flight for each of thse two trajectoires.

The Attempt at a Solution


I tried to do these E= K.(q)/r^2 , E.A.cos (x). well i don't have any clue...

I hope you can help me

Thanks

Welcome to PF.

You can treat it like a traditional projectile range problem.
http://www.physics.ubc.ca/~outreach/phys420/p420_00/darren/web/range/range.html

To figure your acceleration you can use:

[tex] \vec{F} = q * \vec{E} = m * a[/tex]
 
  • #3
Thank you so Much .

http://www.physics.ubc.ca/~outreach/phys420/p420_00/darren/web/range/range.gif

sorry i am not very good in physics , do i need to use this ?
we have x= 1.27mm / so i can get the sin ?
 
  • #4
Gaara09 said:
Thank you so Much .

sorry i am not very good in physics , do i need to use this ?
we have x= 1.27mm / so i can get the sin ?

Once you figure g you can solve for θ.
 
  • #5
g is 10 right ?
 
  • #6
Gaara09 said:
g is 10 right ?

No. g = a. That is the acceleration from your E-field. It is a function of the charge of the proton and its mass. You have to figure that out.

I said you could treat it LIKE a projectile range equation. Only you aren't concerned with gravity. You are concerned with the constant acceleration due to the E-field.
 
  • #7
:cool: aha ok Thanks , got it Now .

Gonna solve it now , Thanks again !
 

Related to How do I use Gauss's Law to calculate proton trajectories in an electric field?

1. What is Guass's law?

Guass's law is a fundamental law in electrodynamics that relates the electric flux through a closed surface to the total charge enclosed by that surface.

2. How does Gauss's law apply to protons?

Gauss's law applies to protons as they have a positive charge and therefore contribute to the electric flux through a closed surface.

3. Can Gauss's law be used to calculate the electric field of a single proton?

No, Gauss's law can only be used to calculate the electric field from a collection of charges, not a single charge.

4. What is the difference between Gauss's law and Coulomb's law?

Gauss's law is a general law that relates the electric field to the charge distribution, while Coulomb's law specifically applies to point charges and directly calculates the electric force between two charges.

5. How does the electric field due to protons change with distance?

The electric field due to protons follows an inverse square law, meaning it decreases with the square of the distance from the proton.

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