How Is Gauss's Law Applied to Calculate Deflection Angle in Charge Interaction?

[Graham1]

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

$$\Phi$$ = ∮EdA

The Attempt at a Solution

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

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

can anybody help me with the method between these two points, or correct me if my method is wrong?

 

[SporadicSmile]

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

$$F = q_1 q_2 / 4 \pi \epsilon_0 r^2$$

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 =]

 

[Graham1]

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 $$r^2=x^2+d^2$$, and that $$x=r tan \theta$$, 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 $$v^2$$ 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?

 

[SporadicSmile]

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 $$\theta$$ is enclosed by that, and from there is simple substitution).

 

[pikorodaimaku]

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.

 

[pikorodaimaku]

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 $$\theta$$ is enclosed by that, and from there is simple substitution).

motherfucker then do it.