Angle of Deflection for a Moving Charge Passing a Fixed Charge

In summary, the conversation discusses a problem involving a moving charged particle and a stationary charge. The particle is initially at x=-infinity and ends up at x=+infinity, while the stationary charge is located at x=0, y=-d. The question is to determine the angle through which the moving charge is deflected as it passes the stationary charge. The conversation also mentions the use of Gauss's law to evaluate an integral in determining the y component of velocity. The attempted solution involves using electric force and F=ma equations, but the calculation of electric field is incorrect. The correct method involves integrating the radial component of electric field over x using Gauss's law.
  • #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}[/tex)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
How did you manage to bring mass into the formula for electric field?
Anyway, your calculation of electric field is wrong, since E is not constant. What you need in this excercise is an integral of Er(radial component of electric field) over x, which can be easily obtained from Gauss law (since the same integral also appears in Gauss law).
 
  • #3
getting mass into the equation was easy enough - use the field E to get a force F.
then simply take F=ma and rearrange
 
  • #4
Physical units are a usefull thing. Especially if they are the same on both sides of the equation.
 

1. What is Gauss Law?

Gauss Law is a fundamental law in electromagnetism that relates the electric field at a given point to the charge enclosed by a closed surface around that point. It is named after the German mathematician and physicist, Carl Friedrich Gauss.

2. How does Gauss Law apply to 2 charges?

Gauss Law can be applied to 2 charges by considering each charge as a point charge and calculating the electric flux through a closed surface surrounding both charges. The total electric flux through the surface is equal to the sum of the electric flux through each individual charge.

3. What is the formula for Gauss Law with 2 charges?

The formula for Gauss Law with 2 charges is given by: ΦE = (q1 + q2) / ε0, where ΦE is the electric flux, q1 and q2 are the charges, and ε0 is the permittivity of free space.

4. How is the direction of electric field determined using Gauss Law?

The direction of the electric field can be determined using Gauss Law by looking at the direction of the electric flux. If the electric flux is entering the closed surface, then the electric field is pointing towards the charges. If the electric flux is exiting the closed surface, then the electric field is pointing away from the charges.

5. Is Gauss Law applicable to all types of charges?

Yes, Gauss Law is applicable to all types of charges, including point charges, distributed charges, and even continuous charge distributions. As long as the charge is enclosed by the closed surface, Gauss Law can be applied to calculate the electric field at a given point.

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