- #1

Heisenberg7

- 59

- 10

- 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?

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?