Gauss Law

1. Jul 9, 2015

erisedk

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 through which the moving charge is deflected.

2. Relevant equations

3. The attempt at a solution
We can't use impulse-momentum or energy equations here. I wrote F = kqQ/d^2 at the origin, dy = 1/2 * kqQ/d^2m * dt^2 where dt is x/v, but I don't know what I'm trying to do, or how I'm going to get there. There's also this thing about considering an infinitely long Gaussian cylinder, but I don't know how that'll fit here.

2. Jul 10, 2015

Nathanael

Just to make sure no confusion arises, you should say y=-D (because lowercase d is used for differentials).

Why are you trying to integrate dy? Obviously the y-coordinate will keep increasing indefinitely even after the particles are infinitely far (so your integral will diverge and be meaningless).

When they say find the angle it is deflected, they want the angle between the initial direction (the +x-axis) and the final direction of motion.

3. Jul 10, 2015

erisedk

Where should I start?

4. Jul 10, 2015

Nathanael

Find the final vertical component of velocity.

5. Jul 10, 2015

erisedk

How? I thought of using v^2 = u^2 + 2as since it's a very small time that the force is gonna act, but then s (i.e. y) comes in.

6. Jul 10, 2015

Nathanael

Use the definition of ay:

dvy=aydt

7. Jul 10, 2015

erisedk

Got it!
I used what it said to use, substituting in Ecosθdx. 2πd in the integral with Q/ε
Then tanθ = vy/vx