# Gauss Law and 2 charges

1. May 27, 2008

### Graham1

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

$$\Phi$$ = EdA

3. 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?

2. May 27, 2008

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, dont 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 =]

3. May 27, 2008

### Graham1

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?

4. May 28, 2008

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 travelled in the x direction and distance travelled in y direction in a given time t, The angle $$\theta$$ is enclosed by that, and from there is simple substitution).

5. Jun 13, 2008

### pikorodaimaku

dude don't you get it? there no Vx, so the problem can't be solved. They should give you Vx. It doesnt matter if you use Vy/Vx=tan(()) or Position Y/x=tan(()) you still need the x component.

6. Jun 13, 2008

### pikorodaimaku

motherfucker then do it.