Potential energy of a system

[flyingpig]

Homework Statement

[PLAIN]http://img716.imageshack.us/img716/1877/35055686.png

The Attempt at a Solution

I am guessing for part (a) I need to find the work it takes to assemble all of those charges together.

I should also set one of the charges as a "reference" from where I meaning my energy from, but since this is a square, I guess it doesn't matter.

[PLAIN]http://img683.imageshack.us/img683/7997/91742091.png

$$\sum U = k\left(\frac{q_1 q_2}{L} + \frac{q_1 q_3}{\sqrt{2}L} + \frac{q_1 q_4}{L} + \frac{q_2 q_3}{L} + \frac{q_2 q_4}{\sqrt{2} L} + \frac{q_4 q_3}{L}\right )$$

$$\sum U = kq^2\left(\frac{4}{L} + \frac{\sqrt{2}}{L}\right )$$

$$\sum U = kq^2\left(\frac{4 + \sqrt{2}}{L} \right )$$

Now for part (b) I am a bit hesitant, should I use the potential energy of the system to find the speed or the individual potential energy with reference to charge Q1?

 

[tedbradly]

The potential energy of the first configuration minus that of the second configuration gives you the energy that has transformed into kinetic energy.

 

[flyingpig]

The potential energy of the first configuration minus that of the second configuration gives you the energy that has transformed into kinetic energy.

What is first and second configuration?

edit: fixed

 

[flyingpig]

Should I find the energy due to the other three charges of one of the particles and then go from there?

 

[SammyS]

f-p, Your potential energy, U, looks good.

KE_total=4·KE_individual.

So I would use the PE of the whole system as well as the KE of the whole system, ie KE_total.

I'm sure he means the second configuration is where all particles have doubled their distances from the center. BTW in this case, just replace L with 2L. Do you see why?

 

[flyingpig]

f-p, Your potential energy, U, looks good.

KE_total=4·KE_individual.

Huh, I was going to re do the addition again... I don't know why I didn't thought of that...

So I would use the PE of the whole system as well as the KE of the whole system, ie KE_total.

I'm sure he means the second configuration is where all particles have doubled their distances from the center. BTW in this case, just replace L with 2L. Do you see why?

I am just going to guess it is because says it doubles...? But shouldn't it be 2√2L then?

Can't I do this $$-\Delta K_{individual}= \frac{1}{4}kq^2\left(\frac{4 + \sqrt{2}}{2\sqrt{2}L} \right )$$

 

[SammyS]

Each charge starts out L/√(2) from the center (that's 1/2 a diagonal). Double the distance from the center, that's 2L/√(2) , so the diagonal is then 4L/√(2) = 2L(√(2)). A side of a square equals the diagonal divided by √(2). Therefore, when the particles have doubled their distance from the center, they also have doubled their distance from each other.

 

[flyingpig]

Each charge starts out L/√(2) from the center (that's 1/2 a diagonal). Double the distance from the center, that's 2L/√(2) , so the diagonal is then 4L/√(2) = 2L(√(2)). A side of a square equals the diagonal divided by √(2). Therefore, when the particles have doubled their distance from the center, they also have doubled their distance from each other.

How did you get this 4L/√(2) = 2L(√(2)) from 2L/√(2).

If you multiply √(2)/√(2) to 2L/√(2) you still get L√(2)

 

[SammyS]

The diagonal is 2 times the distance from the center. i.e. 2·2L/√(2) = 4L/√(2) = 2L(√(2)).

 

[flyingpig]

The diagonal is L/√(2) long originally, twice that we get 2L/√(2)

I don't understand how you got 4

 

[SammyS]

The diagonal is L/√(2) long originally, twice that we get 2L/√(2)

I don't understand how you got 4

The diagonal is L×√(2) long.

The distance from any corner to the center is L/√(2) .

 

[flyingpig]

That aside dividing it by four wouldn't work. The sum of the energy for one charge is

$$\sum U_1 = k\left(\frac{q_1 q_2}{L} + \frac{q_1 q_3}{\sqrt{2}L}+\frac{q_1 q_4}{L}\right)$$

$$\sum U_1 = kq^2\left(\frac{4+\sqrt{2}}{2L} \right )$$

Which is half of the total potential energy of the system

Also okay I understand now how you got it the diagonal thing now, but how did you immediately deduct 2√(2)L to just 2L? I am afraid of carrying on any further because I am unsure why you divided it by 4? The sum of the potential energy due to the other three charge is different than dividing it the sum of the potential energy by 4

 

[SammyS]

...
Also okay I understand now how you got it the diagonal thing now, but how did you immediately deduct 2√(2)L to just 2L? I am afraid of carrying on any further because I am unsure why you divided it by 4? The sum of the potential energy due to the other three charge is different than dividing it the sum of the potential energy by 4

Where did I say to divide by 4?

What quantity did I say should be divided by 4?

 

[flyingpig]

The total energy

 

[flyingpig]

Total energy divided by 4 because there are four masses...?

 

[SammyS]

I never said to divide by 4. You may have interpreted it that way because I said " KE_total = 4·KE_individual ".

Work with the system as a whole and use conservation of energy to find the Kinetic Energy (KE_total) of the system when the particles have doubled their distances from center of the square (i.e., when ALL the mutual distances have doubled).
Then, (Now I say it for the 1st time) divide KE_total by 4 to find KE_individual - KE for each particle.

You are correct in not wanting to divide total energy by 4 ( although it can give the correct answer ) because it makes no sense to divide potential energy, U, of the whole system by 4.

 

[flyingpig]

Now I am really confused should I divide it by four or not?

Should multiply by the system by 4 which will then be the total KE?

 

[flyingpig]

When the distance doubles each U₁, U₂, U₃, U₄ gets

$$U = kq^2 \frac{4 + \sqrt{2}}{4L}}$$

Now this is the energy for one particle, do I need to worry about dividing now? Or can I just set U = K and solve?

 

[flyingpig]

Wait should I let the initial energy of the system

$$kq^2\frac{4 + \sqrt{2}}{L}$$

$$\Delta U = kq^2 \frac{4 + \sqrt{2}}{4L}} - kq^2\frac{4 + \sqrt{2}}{L}$$

And that would be equal to $$-\Delta K$$?

Please help! It's due in like two days

 

[flyingpig]

Wait, forget that one. The sum of energy has nothing to do with the change in energy

$$\Delta U = kq^2\frac{4+\sqrt{2}}{4L} - 0$$ (for the inital energy is 0 since they were infinitely far away)

$$\Delta U = kq^2\frac{4+\sqrt{2}}{4L} = -\Delta K$$

Now I am stuck, I can't take negative square roots.

 

[flyingpig]

Anyone...?

 

[SammyS]

I never said to divide by 4. You may have interpreted it that way because I said " KE_total = 4·KE_individual ".

Work with the system as a whole and use conservation of energy to find the Kinetic Energy (KE_total) of the system when the particles have doubled their distances from center of the square (i.e., when ALL the mutual distances have doubled).
Then, (Now I say it for the 1st time) divide KE_total by 4 to find KE_individual - KE for each particle.

You are correct in not wanting to divide total energy by 4 ( although it can give the correct answer ) because it makes no sense to divide potential energy, U, of the whole system by 4.

Now I am really confused should I divide it by four or not?

Should multiply by the system by 4 which will then be the total KE?

You need to be more specific than "it".

I thought I was very clear. To repeat, "... divide KE_total by 4 to find KE_individual ..." Maybe the hyphen confused you so I omitted it here.

Wait, forget that one. The sum of energy has nothing to do with the change in energy

$$\Delta U = kq^2\frac{4+\sqrt{2}}{4L} - 0$$ (for the initial energy is 0 since they were infinitely far away)

$$\Delta U = kq^2\frac{4+\sqrt{2}}{4L} = -\Delta K$$

Now I am stuck, I can't take negative square roots.

(Yes, you can take negative square roots. What's a problem is taking the square root of a negative number. I.e. ‒√(5) is OK, (so is √(‒x), if x < 0). However, √(‒7) is a problem if you're expecting a real number result. )

Now, to address the main issue here: The initial Kinetic Energy is zero. Therefore, the initial energy (Total Energy by default) is the initial Potential Energy, U.

Your initial PE (of the system) was:
$$\sum U = kq^2\left(\frac{4 + \sqrt{2}}{L} \right )\ .$$

Your final PE (of the system) should be:
$$\sum U = kq^2\left(\frac{4 + \sqrt{2}}{2L} \right )\ .$$
(Found by using 2L instead of L.)

PE_initial+KE_initial = PE_final+KE_final ← This is for the system.

The final KE for the system is given by:

$$\text{KE}=kq^2\left(\frac{4 + \sqrt{2}}{2L} \right )\ .$$

This is the sum of the KEs for all 4 particles. By symmetry, you can infer that each particle has 1/4 of this as its KE when their distance from each other has doubled.

 

[flyingpig]

So we are going to use the energy sum system.

I redid the addition for the total energy after the distance doubles and also got 2L, I still don't know how you managed to deduct it immediately to 2L

Anyways

$$\Delta U = kq^2\frac{4+\sqrt{2}}{2L} - kq^2\frac{4+\sqrt{2}}{L} = -\Delta K$$

$$\Delta U = -kq^2\frac{4+\sqrt{2}}{2L} = -\Delta K$$

Now it looks better because I can take positive square roots now.

$$\Delta U = -kq^2\frac{4+\sqrt{2}}{2L} = -\Delta K$$

$$\Delta U = kq^2\frac{4+\sqrt{2}}{2L} = 4\left ( \frac{1}{2}mv^2 \right )$$

$$|v| =q\sqrt{\frac{k}{2mL}(4+\sqrt{2})}$$

I got it right?

 

[SammyS]

Check your algebra.

I get:
$$|v| =q\sqrt{\frac{k}{4mL}(4+\sqrt{2})}=\frac{q}{2}\,\sqrt{\frac{k}{mL}(4+\sqrt{2})}$$

 

[flyingpig]

Thanks lol...

 

[SammyS]

... I still don't know how you managed to [STRIKE]deduct[/STRIKE] deduce it immediately to 2L ...

It's merely using the idea of a scale factor.

Any linear measurement is multiplied by 2.
Any area measurement is multiplied by 2² = 4.
Any volume measurement is multiplied by 2³ = 8.​

 

[mquirce]

I am confused not with 4 electric charges but even with 2.
Let se: 2 unity charges in a distasnce R = (C / (2*pi*alpha^-1*1) = 338181.8762 m. in static position.
The potential energy of this system is: E = e^2 / (4*pi*epsilon0*r) = h/1 joul.
Now let supose distance = R+1 m. My dilema is:
E < h/1 or E =0 If i admit first then Plank quanta is senseless. If i admit second then interaction of two electric unity charges is not spread to infinity.
I am a curious lay man and with a lame english. Any clue will be wellcome.

 

[flyingpig]

It's merely using the idea of a scale factor.

Any linear measurement is multiplied by 2.
Any area measurement is multiplied by 2² = 4.
Any volume measurement is multiplied by 2³ = 8.​

But area and volume only works for easy ones like squares right?

 

[SammyS]

Hello mquirce.

Please ask your question in a new thread - so that you might get a reply. We were dealing with an introductory physics problem. Your question seems to be very different, having quantum mechanical implications and all.

 

[flyingpig]

I am confused not with 4 electric charges but even with 2.
Let se: 2 unity charges in a distasnce R = (C / (2*pi*alpha^-1*1) = 338181.8762 m. in static position.
The potential energy of this system is: E = e^2 / (4*pi*epsilon0*r) = h/1 joul.
Now let supose distance = R+1 m. My dilema is:
E < h/1 or E =0 If i admit first then Plank quanta is senseless. If i admit second then interaction of two electric unity charges is not spread to infinity.
I am a curious lay man and with a lame english. Any clue will be wellcome.

I don't even know what you did lol... or maybe I am just not smart enough to understand.

Potential energy is a scalar, just add the four particle interactions.

 

[SammyS]

But area and volume only works for easy ones like squares right?

No. If a sphere has a surface area of 7 cm², and a second sphere has a radius twice radius of the first sphere, then the second sphere has an area of 2² times the area of the first sphere, which gives 28 cm².

As long as you scale length, width, and height all the same, the shape of the object doesn't matter .

 

[flyingpig]

This is getting off topic but how can spheres have surface areas? Is it just a circle?

 

[SammyS]

This is getting off topic but how can spheres have surface areas? Is it just a circle?

You're kidding! Think about it.
A sphere is a three dimensional object. A circle is a two dimensional object. Both have a similar definitions.