# Homework Help: Why test charge slows down in terms of electric potential and EPE.

1. Jul 26, 2011

### rcmango

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. Jul 26, 2011

### 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:
$$\frac{1}{4 \pi \varepsilon_0 } \frac{ q_1 q_2 }{r_{12} }$$
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.

3. Jul 26, 2011

### 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.

4. Jul 26, 2011

### BruceW

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"...

5. Jul 27, 2011

### 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.

6. Jul 27, 2011

### 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:
$$\frac{1}{4 \pi \varepsilon_0 } \ \frac{Qq}{r}$$
(where r is the distance between the two charges).
There's also the KE of the small charge $KE_q$ and the KE of the big charge $KE_Q$.
So the important step is to recognise the conservation of energy, that is to say:
$$KE_q \ + \ KE_Q \ + \ EPE \ = \ constant$$
And the big charge is fixed, so its KE is constant, which means $KE_q + EPE = another \ constant$
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

7. Jul 28, 2011

### 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?

8. Jul 28, 2011

### BruceW

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

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.

9. Aug 4, 2011

### rcmango

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:

10. Aug 4, 2011

### 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.

11. Aug 6, 2011

### rcmango

$$V \times q = EPE$$