Gauss Law and 2 charges

In summary, the conversation discusses the determination of the angle through which a moving charge is deflected by a fixed charge. The solution involves using electric force and F=ma equations, calculating the distance of the particle at any given x, and integrating the force with respect to time to find the velocity in the y direction. The angle is then determined by the ratio of the y-velocity to the x-velocity.
  • #1
Graham1
4
0

Homework Statement



A Particle of mass m and chargeq oves at high speed along the x axis. It is initially near x=-infinity and it ends up near x=+infinity. A second charge Q is fixed at the point x=0, y=-d. As the moving charge passes the stationary chrge, its x component o velocity does not change appreciably, but it acquires a small velocity in the y direction. Determine the angle angle through which the moving charge is deflected. suggestion: The integral you encounter in determining v_y can be evaluated by applying Gauss's law to a long cylinder of radius D, centred on the stationary charge.


Homework Equations



[tex]\Phi[/tex] = EdA


The Attempt at a Solution



So Far, I have managed to get to E=Q/2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)
and ended up with a= Qq/(2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)
by using electric force and F=ma equations

I know the answer I should get is [tex]\Theta[/tex]=Qq/(2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)v[tex]^{2}[/tex]

can anybody help me with the method between these two points, or correct me if my method is wrong?
 
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  • #2
you are along the right lines here, but you need to consider that the particle will always experience a force, for all x, not just at x = 0, so work out the distance that the particle is for any x, and use this as your distance.

other than that, don't forget that

[tex]
F = q_1 q_2 / 4 \pi \epsilon_0 r^2
[/tex]

is the force, and from this, and what i said above, you should be able to follow the algebra through and get the right answer =]
 
  • #3
SporadicSmile said:
you are along the right lines here, but you need to consider that the particle will always experience a force, for all x, not just at x = 0, so work out the distance that the particle is for any x, and use this as your distance.

OK, so I'm assuming that I am using [tex]r^2=x^2+d^2[/tex], and that [tex]x=r tan \theta[/tex], which should give me the distance, as I can't see another way of getting it.
but this complicates things on the rhs giving me a "+1 " when simplified, which looks tough to get rid of.

I still can't see a way of getting the [tex] v^2 [/tex] unless I use kinetic energy, but then that seems to cause more problems than it solves.

is it the K.E. I'm looking at, or am I missing something a bit more simple?
 
  • #4
Yes for the first thing, no for the second. The direction r is the direction of the force on the particle, not the direction of the particle itself.

first you want to calculate the velocity in the y direction by integrating the force with respect to time, which is the momentum, and hence you have the velocity.

The angle is calculated from the ratio of the y-velocity to the x-velocity (this can be seen by drawing a triangle, using distance traveled in the x direction and distance traveled in y direction in a given time t, The angle [tex] \theta[/tex] is enclosed by that, and from there is simple substitution).
 
  • #5
dude don't you get it? there no Vx, so the problem can't be solved. They should give you Vx. It doesn't matter if you use Vy/Vx=tan(()) or Position Y/x=tan(()) you still need the x component.
 
  • #6
SporadicSmile said:
Yes for the first thing, no for the second. The direction r is the direction of the force on the particle, not the direction of the particle itself.

first you want to calculate the velocity in the y direction by integrating the force with respect to time, which is the momentum, and hence you have the velocity.

The angle is calculated from the ratio of the y-velocity to the x-velocity (this can be seen by drawing a triangle, using distance traveled in the x direction and distance traveled in y direction in a given time t, The angle [tex] \theta[/tex] is enclosed by that, and from there is simple substitution).

motherfucker then do it.
 

1. What is Gauss Law?

Gauss Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed by that surface.

2. How does Gauss Law apply to 2 charges?

Gauss Law can be used to calculate the electric field between and around 2 charges by considering the net flux through a closed surface surrounding both charges.

3. What is the equation for Gauss Law?

The equation for Gauss Law is ∮E · dA = Qenc/ε0, where ∮E · dA represents the electric flux through a closed surface, Qenc is the net charge enclosed by the surface, and ε0 is the permittivity of free space.

4. Can Gauss Law be used for any distribution of charges?

Yes, Gauss Law can be used for any distribution of charges as long as the net charge enclosed by the surface is known.

5. What is the significance of Gauss Law in electromagnetism?

Gauss Law is a crucial tool in understanding the behavior of electric fields and charges, and it is used in various applications such as designing electronic devices and analyzing electric circuits.

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