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Oscillating charges in an electric field

  • Thread starter WrongMan
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  • #1
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I dont have a single problem, my teachers use this a lot.
a few examples, are an uniformely charged circle and a point charge in the middle, where if you move the point charge a small distance perpendicular to the circle, it would begin to oscilate, and i need to find the frequency of oscilation.
or, just a couple of point charges on the x-axis at a certain distance to the origin, and you place an oposite charge on the origin and move it a small ammount on the y axis, what would be the frequency of oscilation, or find an expression for the position of the oscilating charge.

equations i think would be relevant are electric force and harmonic oscilation:
Fe = kqQ/r^2 and F=-Kx
Would the answer be just to substitute the forces for eachother and have
kqQ/r^2 = -Kx
Or do i have to find K for each specific situation?
these are "old" problems as in i'm not studying this part of the program anymore, Electric fields and forces are gonna be in the final, and im pretty sure a problem like this is gonna show up.
I havent tried to solve the exercises since i dont have the answers i have no idea to check if im correct other than criticizing my results
Im sorry i'm not showing any real attempt at solving these, i just need a little push in the right direction on how do i have to think about this?
 

Answers and Replies

  • #2
cnh1995
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are an uniformely charged circle and a point charge in the middle, where if you move the point charge a small distance perpendicular to the circle, it would begin to oscilate, and i need to find the frequency of oscilation.
You need to first find a general expression for electric field E(x) on the axis, at a distance x from the center of the ring. Then, using F(x)=q*E(x), you can get the force on the charge. For simple harmonic motion, the displacement 'x' of the charge should be small enough compared to the radius 'r' of the ring, such that F=kx. Hence, assuming x<<r in the expression for F(x), you'll get the expression for force of SHM in the form of F=kx.
 
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  • #3
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You need to first find a general expression for electric field E(x) on the axis, at a distance x from the center of the ring. Then, using F(x)=q*E(x), you can get the force on the charge. For simple harmonic motion, the displacement 'x' of the charge should be small enough compared to the radius 'r' of the ring, such that F=kx. Hence, assuming x<<r in the expression for F(x), you'll get the expression for force of SHM in the form of F=kx.
Ok great that's what i had in mind. thanks!
 
  • #4
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Ok so i returned to this today.
The exercise i looked at was the following:
there are two charges (q>0) positioned at (+a,0) and (-a,0)
another test charge (same signal) is placed at the origin, and can only move along the x axis, find the oscilation frequency when you give it a small displacement along the x axis.

So i found the field expression for both original charges. zeroed the y component and got
for the charge at -a:
E=kq*(1/(x+a)^2) * (x+a)/((x+a)^2)^1/2
for the one at +a
E=kq*(1/(x-a)^2) * (x-a)/((x-a)^2)^1/2

Found the resulting force on the test charge that was displaced dx
Ft=kqqtest*(1/(dx+a)^2 - 1/(dx-a)^2)
then i use this force to replace in "F=-kdx".
solve for k, then since (k/m)^1/2 = 2pi/T, solve for T then T = 1/f solve for f.
is this correct?
 
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  • #5
cnh1995
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there are two charges (q>0) positiones at (+a,0) and (-a,0)
If both the charges have same sign, y compoment of the field is not 0.
 
  • #6
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If both the charges have same sign, y compoment of the field is not 0.
yes, they have the same charge and are positioned along the x axis, so on the x axis the field in the y direction is 0 (even if they had opposite charges it would be 0).

I'm considering only the X-axis on this part of the problem i meant that previously i found the field at any position on the XY plane, got:
E=kq*(1/((x+a)^2+y^2) * (x+a,y)/((x+a)^2+y^2)^1/2
for the one at +a
E=kq*(1/((x-a)^2+y^2) * (x-a,y)/((x-a)^2+y^2)^1/2

but since for this part of the problem the test charge in only in the x axis, i zeroed the y's in those expressions, thats what i meant.
 
  • #7
cnh1995
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so on the x axis the field in the y directions is 0 (even if they had opposite charges it would be 0)
Oh..I think I misread your question. On the x axis, the y component of the field is zero but elsewhere, it is non-zero. It would be true for opoosite charges as well. I pictured that the test charge is displaced along y axis and got confused.
 
  • #8
ehild
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Found the resulting force on the test charge that was displaced dx
Ft=kqqtest*(1/(dx+a)^2 - 1/(dx-a)^2)
then i use this force to replace in "F=-kdx".
solve for k, then since (k/m)^1/2 = 2pi/T, solve for T then T = 1/f solve for f.
is this correct?
Ft is not proportional to dx. What do you mean with replacing F in F=-kdx with it? Show how do you solve for k, please.
 
  • #9
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Ft is not proportional to dx. What do you mean with replacing F in F=-kdx with it? Show how do you solve for k, please.
what? how is it not proportional? Ft must be proportional to dx, it makes sense that as dx aproaches (a,0), force on the test charge by the charge in (a,0) goes to infinity, and the force on the test charge by the charge on (-a,0) goes to force at a distance of 2a.

-(k/dx)*qqtest*(1/(dx+a)^2 - 1/(dx-a)^2)=k
the minus makes sense since i moved the charge a positive dx, the force would have "-x direction"
this and the rest is just substitution.
Am i wrong?
 
  • #10
ehild
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what? how is it not proportional? Ft must be proportional to dx, it makes sense that as dx aproaches (a,0), force on the test charge by the charge in (a,0) goes to infinity, and the force on the test charge by the charge on (-a,0) goes to force at a distance of 2a.
That does not mean a force proportional to dx.
-(k/dx)*qqtest*(1/(dx+a)^2 - 1/(dx-a)^2)=k
the minus makes sense since i moved the charge a positive dx, the force would have "-x direction"
this and the rest is just substitution.
Am i wrong?
Yes.
First: you use the symbol k for two different things, the constant in Coulomb's law, and the force constant in F=-kx. k just cancels from the equation you presented.
What do you get for the force constant from your equation?
 
  • #11
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That does not mean a force proportional to dx.

Yes.
First: you use the symbol k for two different things, the constant in Coulomb's law, and the force constant in F=-kx. k just cancels from the equation you presented.
What do you get for the force constant from your equation?
oh ok i get it, i was saying it was proportional as in Ft varies when dx changes not direct or inverse proportionality.

oh yeas i didnt notice i used the same letter for two different things.
the force constant would be:
k1= -(k2/dx)*qqtest*(1/(dx+a)^2 - 1/(dx-a)^2)

where k1 is the force constant and k2 is the constant in coulomb law
 
  • #12
ehild
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the force constant would be:
k1= -(k2/dx)*qqtest*(1/(dx+a)^2 - 1/(dx-a)^2)

where k1 is the force constant and k2 is the constant in coulomb law
Yes, but the force constant should be a constant. Yours depends on dx. You have to simplify the expression for k1 and use that dx<<a.
Hint: bring the terms of 1/(dx+a)^2 - 1/(dx-a)^2 to common denominator, expand the squares and simplify, ignore terms with dx on power higher than 1.
 
  • #13
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Yes, but the force constant should be a constant. Yours depends on dx. You have to simplify the expression for k1 and use that dx<<a.
Hint: bring the terms of 1/(dx+a)^2 - 1/(dx-a)^2 to common denominator, expand the squares and simplify, ignore terms with dx on power higher than 1.
yeah makes sense, i got
K1=(4qqtestk2)/a3
 

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