Why test charge slows down in terms of electric potential and EPE.

[rcmango]

Homework Statement

a large positive charge +Q fixed at some location in otherwise empty space, far from all other charges. A positive test charge of smaller magnitude +q is launched directly towards the fixed charge. Of course, as the test charge gets closer, the repulsive force exerted on it by the fixed charge slows it down. Your job is to explain why the test charge slows down, but in terms of electric potential and EPE, rather than in terms of fields or forces.

explain how the electrical potential encountered by the test charge changes as it gets closer to the fixed charge, and why.

explain how the EPE of the test charge changes as it gets closer to the fixed charge, and why.

Homework Equations

The Attempt at a Solution

The EPE difference between q and Q does not change, it says the same proven using this formula k(Q-q)/r, the EPE is always increasing as the test charge approaches the fixed charge, q EPE is increasing and Q's EPE is increasing.

The work done by the net E field in moving a test charge a distance defines EPE. So's as q moves closer to Q the fixed charge, q gains PE, in turn q also gains EPE. The voltage is increasing.
The voltage from the equipotential fields are not the same due to the fixed charge at the moving charge, or at the moving charge due to the fixed charge. However, the difference due to both charges are the same. ( V = k(Q-q)/r )

The particle is slowing down and eventually stops when the potentials are equal amounts.

below, is a response to my paragraph, I am really having a hard time addressing what is going on using PE and EPE, can someone stick with me on this and guide me through writing a short paragraph explaining what is needed to answer the two sentences above. Thanks alot.

The expression k(Q-q)/r doesn't apply to this situation. The test charge q is simply experiencing the potential created by the fixed charge Q. You say that q gains PE and therefore also EPE, but the only kind of PE that q gains is EPE. Start with what happens to the potential as q gets closer to Q, then deduce what happens to its EPE, and then use the energy conservation principle to explain why the charge slows down.

 

[BruceW]

Electrical potential energy is a type of potential energy. You seem to be talking about them separately. But because the only type of PE is EPE (in this situation), this means EPE and PE are the same thing.

The electrostatic potential energy of two point charges is:
[tex] \frac{1}{4 \pi \varepsilon_0 } \frac{ q_1 q_2 }{r_{12} } [/tex]
It makes no sense to talk about the EPE of one and the EPE of the other. The equation above gives the total EPE of both.

 

[hnought]

The two particles form a closed system and hence the total energy must remain the same. hence: initial kinetic energy + initial potential energy = final kinetic energy + final potential energy.

The initial potential energy (at least at infinity) can be taken as zero and the final kinetic energy (where q stops) is also zero. Therefor, the initial kinetic energy given by classical mechanics is 1/2 m(v^2). This must equal (kQq)/r. Knowing r (distance from Q where q stops and mass of q for instance will allow you to calculate initial velocity, etc, etc.

J.

 

[BruceW]

The EPE difference between q and Q does not change, it says the same proven using this formula k(Q-q)/r, the EPE is always increasing as the test charge approaches the fixed charge, q EPE is increasing and Q's EPE is increasing.

The EPE of the system does increase as the small charge gets closer to the big charge. But I don't know what your formula represents, and I don't know what you mean by "the EPE difference between q and Q does not change"...

 

[rcmango]

"explain how the electrical potential encountered by the test charge changes as it gets closer to the fixed charge, and why."

...Okay, I can see that my first statement in my original post is confusing, I want to start answering the question I have in quotes above. So where do we start?

I understand now that this problem is a closed system with two particles. One is fixed and one is moving toward the other one. Also, I should be looking at the Potential energy aka the Electrical Potential Energy as a whole, not as a separate dynamic. Please help me answer this question.

 

[BruceW]

Yep. The question also asked to explain the problem in terms of energy.
So to start with, we need to think about all the forms of energy in the question.
There's the EPE of the system, which is given by:
[tex] \frac{1}{4 \pi \varepsilon_0 } \ \frac{Qq}{r} [/tex]
(where r is the distance between the two charges).
There's also the KE of the small charge [itex]KE_q[/itex] and the KE of the big charge [itex]KE_Q[/itex].
So the important step is to recognise the conservation of energy, that is to say:
[tex] KE_q \ + \ KE_Q \ + \ EPE \ = \ constant [/tex]
And the big charge is fixed, so its KE is constant, which means [itex] KE_q + EPE = another \ constant [/itex]
This equation is all you need to explain the behaviour of the small charge. So, according to this equation, the change in KE of the small charge is determined by the change in EPE.
To answer the problem, you need to explain how the EPE changes as the small charge gets closer to the big charge. And then you can explain that energy conservation means that a change in EPE results in a change in KE of the small charge

 

[rcmango]

Okay, so far I can say, one is fixed and one is moving toward the other one. Looking at the Potential energy aka the Electrical Potential Energy as a whole. The EPE changes as the small charge approaches the big charge, this in turn changes the KE of the small charge also. Now, how can I determine how the EPE is changing? I am going to say that the EPE is getting greater as the small charge approaches the big charge. Also, when talking about PE which is said to be the same thing, PE is decided upon the EPE and the unit of the charge?

 

[BruceW]

That's a pretty darn good description. It looks like you understand it.

Also, when talking about PE which is said to be the same thing, PE is decided upon the EPE and the unit of the charge?

In a general situation, you would have: PE = GPE + EPE + other potential energies.
But in our situation, the only potential energy we are interested in is the electrical potential energy, so we have: PE = EPE.

 

[rcmango]

Okay, so i made a response and I have a counter response. Please help me put together a final response.

Here is what I said,
In this particular situation, PE is EPE, essentially the same thing. So using equations of the electrostatic potential energy of two point charges is: (1/4pi*e0)(q1q2/r12) So EPE is joined as one. This is a closed system, so the EPE increases as the smaller charge approaches the fixed charge.. Looking at the Potential energy aka the Electrical Potential Energy as a whole. The EPE changes as the small charge approaches the big charge, this in turn changes the KE of the small charge also. the EPE is getting greater as the small charge approaches the big charge. The only potential energy we are interested in is the electrical potential energy, so PE = EPE.

Now here is the counter response:

The simplest way to tackle this is to completely avoid equations and explain in terms of concepts.

The fixed charge Q creates electric potential in the space surrounding it: 50 volts here, 100 volts there, 5000 volts somewhere else. Which of these locations is farthest from Q? Which is closest?

Thinking about this will help you figure out what how the potential at locations closer to Q differs from the potential at locations farther from Q. This is the potential the test charge moves through: does it rise or fall as the test charge q gets closer to Q?

Once you've established how the potential V changes, you can explain what happens to the EPE of the test charge as it gets closer: does it increase or decrease?

Once you know how the EPE of the test charge changes, you can explain why it slows down, using the energy conservation principle.

 

[BruceW]

Yep. What you said and the counter-response are both correct. I think the last step is to say that since the EPE increases as the small charge gets closer, its KE must decrease, to keep the total energy constant.

 

[rcmango]

Okay, how about this response, please take a look:

As the smaller charge gets closer to the larger charge, the closer the smaller charge is to Q the higher the voltages. So 5000 volts would be very close to the larger charge, as well as 50 volts being the furthest away. So as the test charge q gets closer to Q, the potential rises. So as the potential rises, as the test charge gets closer, the EPE, it also rises. As the EPE increases the kinetic energy decreases for the test charge and it begins to slow down, the energy is never lost, it is transferred.

 

[BruceW]

Yep, that all sounds good. One other thing: the potential (a.k.a. 'voltage') is defined as:
[tex] V \times q = EPE [/tex]
So potential at some point in space times the charge at that point is potential energy.
These two terms sound very similar, which I often mixed up when I was first learning.