Coloumb Force

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Charles Darwin

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Two particles of opposite charge are moved slowly apart. As the particles are moved apart, the force between them is:

A. attractive and increasing
B. attractive and decreasing
C. repulsive and increasing
D. repulsive and decreasing

Answer: B

My answer: A

Reasoning: The forces are obviously attracting because they're unlike charges; however, if you use the Coloumb equation F = k (q1q2/r^2), the oppositely charged particles would yield F = – k (q1q2/r^2). Keeping everything else constant besides radius, F = – (1/r^2). If you increase the radius (pull the particles apart), the force between the particles becomes a successively smaller NEGATIVE number, corresponding to an increase in force.

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I see what you are trying to say but signs for forces are used to designate direction only. When talking about increasing/decreasing force, you can safely assume that the question is actually talking about increasing/decreasing magnitude of the force.

Generally, talking about increasing/decreasing vector quantities is poor terminology since vectors are not ordered, only their magnitudes are.

It is likely that you may see some questions about velocity/acceleration which actually violate that and do ask you to compare positive/negative velocities/acceleration. That's rather unfortunate and will happen mostly for single dimension cases.
 
yea it's talking about magnitude. a negative is equivalent to -1 which doesnt do anything to the magnitude and in physics negative numbers just imply direction.
 
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yea it's talking about magnitude. a negative is equivalent to -1 which doesnt do anything to the magnitude and in physics negative numbers just imply direction.

That's correct for vector quantities (like velocities, forces, etc). Negative number have their usual meaning for scalars like energy.
 
Two particles of opposite charge are moved slowly apart. As the particles are moved apart, the force between them is:

A. attractive and increasing
B. attractive and decreasing
C. repulsive and increasing
D. repulsive and decreasing

Answer: B

My answer: A

Reasoning: The forces are obviously attracting because they're unlike charges; however, if you use the Coloumb equation F = k (q1q2/r^2), the oppositely charged particles would yield F = – k (q1q2/r^2). Keeping everything else constant besides radius, F = – (1/r^2). If you increase the radius (pull the particles apart), the force between the particles becomes a successively smaller NEGATIVE number, corresponding to an increase in force.

I think a good learning point can be made here. Lets try this question but talk about the POTENTIAL energy change. I'm going to try to apply what I know and you guys can correct me.

So if you have the same case where the two oppositly charged particles get seperated, their potential energy will decrease right because you have to do work to overcome their attractiveness.


Now if you have two charges that repel each other and you seperate them, the potential energy would increase because there is no work that needs to be done to separate them?
 
I think a good learning point can be made here. Lets try this question but talk about the POTENTIAL energy change. I'm going to try to apply what I know and you guys can correct me.

So if you have the same case where the two oppositly charged particles get seperated, their potential energy will decrease right because you have to do work to overcome their attractiveness.


Now if you have two charges that repel each other and you seperate them, the potential energy would increase because there is no work that needs to be done to separate them?
I'm fairly certain it's the exact opposite. I used your intuition and applied it to this problem which is why I got the question wrong. :laugh:

Two oppositely charged particles have a negative potential energy. If you pull them apart, you're increasing their potential energy because you have to do work to separate them (i.e., Coloumb's Eq. - negative number approaching zero = increasing PE). Likewise, if you have two similarly charged particles and pull them apart you're decreasing the potential energy.

Right?
 
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That's correct for vector quantities (like velocities, forces, etc). Negative number have their usual meaning for scalars like energy.
For example, planets have massive negative potential energies that keep the planets in their orbit (over-simplified, I know); the potential energies must be large and negative because the kinetic energies of the planets cannot exceed the total energy of the system or the planets would fly out of their orbits.
 
I'm fairly certain it's the exact opposite. I used your intuition and applied it to this problem which is why I got the question wrong. :laugh:

Two oppositely charged particles have a negative potential energy. If you pull them apart, you're increasing their potential energy because you have to do work to separate them (i.e., Coloumb's Eq. - decreasing a negative number = increasing PE). Likewise, if you have two similarly charged particles and pull them apart you're decreasing the potential energy.

Right?

Yes I think you are right actually. It would make sense...work=PE. The charges would want to come together "more" the farther they are apart.
 
I'm fairly certain it's the exact opposite. I used your intuition and applied it to this problem which is why I got the question wrong. :laugh:

Two oppositely charged particles have a negative potential energy. If you pull them apart, you're increasing their potential energy because you have to do work to separate them (i.e., Coloumb's Eq. - negative number approaching zero = increasing PE). Likewise, if you have two similarly charged particles and pull them apart you're decreasing the potential energy.

Right?

You are correct.
 
For example, planets have massive negative potential energies that keep the planets in their orbit (over-simplified, I know); the potential energies must be large and negative because the kinetic energies of the planets cannot exceed the total energy of the system or the planets would fly out of their orbits.

Potential energies are tricky. You can never measure a potential energy as an absolute value - that would mean that you're able to account for all the other bodies affecting your experiment and that's just not realistic. Lucky for us, the amount of potential energy does not affect how the body moves - the only thing that matters is how potential energy changes with these movements Because of this you can set the 0 for the potential energy at any arbitrary position. When talking about potential energy due to gravity in planetary scale, it is very convenient to set the 0 to the amount of potential energy that the bodies have when they are infinitely far away from each other.

Here is why: if you consider the potential energy due to gravity between two planets, you'll find that it does have a minimum value when the planets are as close as possible to each other and that it increases asymptotically towards a single value when the planets get infinitely far away from each other. That value will be different for differently sized planets. If you set this value to 0 (like we normally do) you can discard the PE from far away bodies as 0 and focus only on the interaction between the two nears bodies.

You could have just as easily set the 0 to be the minimum of the PE, when the bodies are next to each other. Now when you try to describe your system, you have to consider the large amounts of PE due to remote planets. That just puts arbitrary large numbers in your equations without any practical effect. When you consider the distance to these far bodies, the changes in PE are still going to be insignificant but you still need to carry the large numbers around.
 
That 0 is actually the highest potential energy that the particles are going to have. Bringing them together from that distance will still generate energy.
But once you bring the particles closer than their mutual binding energy, won't the potential energy be higher than if the particles were infinitely far apart? In other words, the potential energy would be highest when the nuclei are repelling one another, as opposed to when the particles are extremely far apart. Or am I misinterpreting the graph?
 
But once you bring the particles closer than their mutual binding energy, won't the potential energy be higher than if the particles were infinitely far apart? In other words, the potential energy would be highest when the nuclei are repelling one another, as opposed to when the particles are extremely far apart. Or am I misinterpreting the graph?

Yes but not because of the Coulomb force. The graph is a combination of the interactions due to two separate forces - the Coulomb force which is pulling the particles together and a strong force, keeping them apart.

If you are talking strictly about Coulomb force, the potential energy will continue to decrease towards negative infinity as the particles get closer. In reality, for small particles, you have strong force interactions which keep the PE from dropping down like that. For large particles, as the distance decreases they eventually get close enough for the charge to cross between them, again limiting how low the potential energy can be.
 
Yes but not because of the Coulomb force. The graph is a combination of the interactions due to two separate forces - the Coulomb force which is pulling the particles together and a strong force, keeping them apart.

If you are talking strictly about Coulomb force, the potential energy will continue to decrease towards negative infinity as the particles get closer. In reality, for small particles, you have strong force interactions which keep the PE from dropping down like that. For large particles, as the distance decreases they eventually get close enough for the charge to cross between them, again limiting how low the potential energy can be.
Ahh, I see. Very well then. 👍
 

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