Find Velocity of Particle for System of Charges

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Klaus von Faust
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Homework Statement


Two charges, ##-q_1## and ##q_2## are fixed in the vacuum and separated by a distance ##a##. What should be the velocity ##v## of a particle with mass ##m## and charge ##q##, traveling from an infinitely far point along the line which unites ##q_1## and ##q_2## in order to get in the point where ##q_1## is located?

Homework Equations


##W=q\Delta\phi##
##T=\frac {mv^2} 2##
##\phi=\frac {q} {4\pi \epsilon_0 r}##

The Attempt at a Solution


I tried to write energy conservation, but the potential interaction energy between ##q_1## and ##q## is infinite because the distance is zero. The velocity has to be zero if ##q_2## is less than ##q_1##
 
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I suppose you have to make some assumptions here, and I think you do the right thing for the first one, assuming ##q_1## is negative and ##q_2## and ##q## are positive. The second is that the approach is from the side of ##-q_1##, because ##q## won't get past ##q2## on this connecting line.

I agree that
Klaus von Faust said:
The velocity has to be zero if ##|q_2|## is less than ##|q1|##
(but perhaps the exercise composer would want you to explain?)

So you are left with the case ##|q_2|> |q1|## and I think here you can leave out wondering aabout
Klaus von Faust said:
potential interaction energy between ##q_1## and ##q##
(they will annihilate) and instead calculate what speed is needed to come sufficiently in the neighborhood of ##q_1##
Make a sketch of the potential along the connecting line.
 
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BvU said:
I suppose you have to make some assumptions here, and I think you do the right thing for the first one, assuming ##q_1## is negative and ##q_2## and ##q## are positive. The second is that the approach is from the side of ##-q_1##, because ##q## won't get past ##q2## on this connecting line.

I agree that
(but perhaps the exercise composer would want you to explain?)

So you are left with the case ##|q_2|> |q1|## and I think here you can leave out wondering aabout
(they will annihilate) and instead calculate what speed is needed to come sufficiently in the neighborhood of ##q_1##
Make a sketch of the potential along the connecting line.
I suppose that the initial potential energy has to be just the interaction energy between ##q_1## and ##q_2##, ##-\frac {q_1 q_2} {4\pi\epsilon_0a}##, because the ##q## point of charge is far away. Then, the kinetic energy of the point of charge transforms completely into potential energy.
##\frac {mv^2} 2=\frac {q q_2} {4\pi\epsilon_0a}## Am I right?
 
BvU said:
No. Did you make the sketch ?
BvU said:
No. Did you make the sketch ?
52161712_2022844134503197_8288425748927086592_n.jpg
 
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BvU said:
Make a sketch of the potential along the connecting line
 
I think I don't understand what do you mean. Do you mean an electric field sketch? Because the potential is a scalar and I don't know how to represent it graphically
 
can you make a sketch of the scalar potential 1/|x| ?
 
The function of the potential in terms of x is ##\phi=\frac {-k q_1} x +\frac {k q_2} {x+a}##
I can not draw the graph of this function, but the graph of the simple potential is something like this
52830823_902735246784573_5704433089892057088_n.jpg
 
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upload_2019-2-19_17-11-36.png


This is a sketch of 5 / |x-5| - 1 / |x+5|
 
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Perhaps a bigger hint. Thinking in terms of forces:

When ##q## is far away, the force from ##q_2## dominates, so the force is repulsive. As ##q## gets closer, the distance squared term eventually balances the forces; and, after that, the attractive force from ##q_1## must dominate.

It's interesting that in the OP you said:

Klaus von Faust said:
The velocity has to be zero if ##q_2## is less than ##q_1##

But, you never really explained that in terms of forces or potential.
 
Klaus von Faust said:
The function of the potential in terms of x is ##\phi=\frac {-k q_1} x +\frac {k q_2} {x+a}##
I can not draw the graph of this function, but the graph of the simple potential is something like this

If you have any function in this case, you must look at ##x \rightarrow 0##, ##x \rightarrow \infty## and look for turning points.

It's doesn't matter how advanced the physics gets, basic calculus is always useful!
 
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PeroK said:
If you have any function in this case, you must look at ##x \rightarrow 0##, ##x \rightarrow \infty## and look for turning points.

It's doesn't matter how advanced the physics gets, basic calculus is always useful!
Thank you very much, now I understood. I just need to find the ##x## at which the forces generated by ##q_1## and ##q_2## on ##q## will be equal in magnitude, this is the boundary condition. And then I apply the conservation of energy and find out the velocity I am seeking for.
 
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Klaus von Faust said:
at which the forces generated
In other words: where the potential has a maximum
 
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