# A charge q approaching two stationary charges q1 and q2

• Heisenberg7
Heisenberg7
Homework Statement
Two charges ##-q_1## and ##+q_2## are stationed at a distance ##l##. A particle of mass ##m## and charge ##q## is approaching the 2 charges along the flat plane which passes through them (photo). What's the minimum velocity the particle must have so that it makes it to the charge ##-q_1##?
Relevant Equations
##\Delta E = 0##
I'm only confused about one part of this problem and that is setting up the conservation of energy equation. In the solution, they just wrote this: $$\frac{mv_o^2}{2} = - k \frac{q_1 q}{r} + k \frac{q_2 q}{l + r}$$ where ##r## represents the distance at which the force created by the negative charge is equal to the force created by the positive charge (assuming that ##\mid q_1\mid < \mid q_2\mid##). Now, I tried doing this my way and I ran into some issues. $$\Delta E = 0 \implies \Delta K + \Delta U = 0$$ Now, this is the part that confuses me. To get the equation from above I have to assume that the magnitude of velocity and distance ##r## is equal to 0 and that the potential energy at some initial point is 0. Is that true? I mean, for the velocity, sure, it could be that the particle isn't fully in an equilibrium so I guess we can assume that the velocity at the distance ##r## can be 0 because even slightest movement could cause the particle to move. Now for the potential energy, does it assume here that the particle is at infinity because how could the potential energy be 0 unless the particle is at infinity?
From this we get, $$(0 - \frac{mv_o^2}{2}) + (-k\frac{q_1 q}{r} + k \frac{q_2 q}{l + r} - 0) = 0 \implies \frac{mv_o^2}{2} = - k \frac{q_1 q}{r} + k \frac{q_2 q}{l + r}$$ Is this it?

Clearly the velocity required is less the closer the point at which it starts with that velocity. Therefore the question ought to have stated that it starts from a distance >> l.

haruspex said:
Clearly the velocity required is less the closer the point at which it starts with that velocity. Therefore the question ought to have stated that it starts from a distance >> l.
Well, it doesn't explicitly say that. It just says that they start far from each other.

Heisenberg7 said:
Well, it doesn't explicitly say that. It just says that they start far from each other.
It is the same thing.

"Far from each other" is taken to mean "far enough so that you don't have to worry about how far".

jbriggs444 said:
It is the same thing.

"Far from each other" is taken to mean "far enough so that you don't have to worry about how far".
So, does that mean that we can assume that the potential energy at that "far enough" is 0?

Heisenberg7 said:
So, does that mean that we can assume that the potential energy at that "far enough" is 0?
Yes. If told that ##q## is 'far away' (or words to that effect) from all other charges, it is implied that ##q##'s potential energy can be taken to be zero. It's fairly common terminology.

A minor point...

In an extreme case where ##|q_1|## and ##|q_2|## are nearly equal (with ##|q_2| \gt |q_1|##) and where ##l## is very small, then ##r## could be very large.

So maybe it's preferable to say that ##q## needs to start from a distance ##\gg r##, rather than from a distance ##\gg l##. But 'far way' is simpler terminology.

Heisenberg7
Heisenberg7 said:
To get the equation from above I have to assume that the magnitude of velocity and distance ##r## is equal to 0
at distance ##r##* For some reason, I can't edit the post.

Heisenberg7 said:
So, does that mean that we can assume that the potential energy at that "far enough" is 0?
I would quibble a bit about that.

Potential energy is always relative to some arbitrary reference level. Only potential differences are physically meaningful.

I do agree that if you are taking zero potential to be "at infinity" then yes, we can assume that "far enough" is potential energy [insignificantly different from] zero.

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