Gauss Law and 2 charges

  • Thread starter Graham1
  • Start date
  • #1
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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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
249
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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
4
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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
249
0
Physical units are a usefull thing. Especially if they are the same on both sides of the equation.
 

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