Calculating the electric potential engery of the system

[mr_coffee]

Hello everyone I'm having problems figuring this out, In the quark model of fundamental particles, a proton is composed of three quarks: two "up" quarks, each having charge +2e/3, and one "down" quark, having charge -e/3. Suppose that the three quarks are equidistant from one another. Take the distance to be 1.32 10-15 m.

(a) Calculate the electric potential energy of the system of only the two "up" quarks.
? eV
(b) Repeat for all three quarks.
? eV

Well i used the following: U_12 = [k*q1*q2]/d = [9.0e9*(2/3)^2*e]/1.32e-15 = 3.0275e24 which is wrong, any ideas what I'm doing wrong?

 

[quantumdude]

U_12 = [k*q1*q2]/d = [9.0e9*(2/3)^2*e]/1.32e-15 = 3.0275e24 which is wrong, any ideas what I'm doing wrong?

You forgot to square the "e".

 

[mr_coffee]

Thanks! I don't see how that effects my answer though, because the answer they want is it eV, so how would i get rid of an e, if its now e^2?

 

[quantumdude]

But you were calculating in SI units, right? That means that $e=1.602\times10^{-19}C$. If you forget to square it, then your answer will be off by 19 orders of magnitude!

 

[mr_coffee]

ohhhh! i didn't nkow that's what the problem was telling me! infact that makes no sense, how can an electron be positive? I tried it anways, and i got it wrong again, maybe i misunderstood whta you ment. I did the following: U_12 = [k*q1*q2]/d = [9.0e9*((2/3)*1.602e-19)^2]/1.32e-15 = 7.77e-14 eV

 

[quantumdude]

infact that makes no sense, how can an electron be positive?

The charge on the electron is not positive. It is $-e$.

I did the following: U_12 = [k*q1*q2]/d = [9.0e9*((2/3)*1.602e-19)^2]/1.32e-15 = 7.77e-14 eV

Examine the units of the left hand side. They are all SI units. That means that your answer will have the SI unit of energy, which is not eV.

 

[mr_coffee]

well he wants the answer in eV, that's what I'm confused on, how would i convert an answer from Joules to eV? id on't even know what eV is. I think he is mnaking it up as he goes along hah

 

[quantumdude]

well he wants the answer in eV,

Nevertheless, if you use all SI units in the formula for electric potential energy, your answer will come out with the SI units of energy. That much you should know.

thats what I'm confused on, how would i convert an answer from Joules to eV? id on't even know what eV is.

The definition of an electron volt is as follows:

1 electron volt (eV) is defined as the kinetic energy acquired by an electron when it is accelerated through a potential difference of 1 Volt.

That definition is sufficient for you to work out the conversion factor.

I think he is mnaking it up as he goes along hah

Nope, this is all in your book, which you should be reading.

 

[mr_coffee]

Calculate Electric potential engery of the system

Hello everyone, i asked the professor and he told me what to do but I'm still getting the wrong answer for some reason.

In the quark model of fundamental particles, a proton is composed of three quarks: two "up" quarks, each having charge +2e/3, and one "down" quark, having charge -e/3. Suppose that the three quarks are equidistant from one another. Take the distance to be 1.32 10-15 m.
(a) Calculate the electric potential energy of the system of only the two "up" quarks.
eV
(b) Repeat for all three quarks.
eV

So I used the equation:
V = (k)(Q/r + Q/r)
V = (9.0E9)* [(2*1.6E-19)]/1.32e-15m + (2*1.6E-19)]/1.32E-15m)
V = 1454545.455 V
So i need to convert to an eV so i divided by e which is 1.6E-19 as my professor said and other people on the forum and i got:
V = 9.0909E24 eV which is wrong! any ideas? Thanks.

 

[Doc Al]

Is that supposed to be work for (a) or (b)? Either way, consider this: When finding the energy for two quarks there is only one pair, and thus only one use of $V = k Q/r$. But when you add the third quark, how many pairs do you end up with?

 

[quantumdude]

mr_coffee:

I've merged your two threads on this question. In the future, please continue with the thread that you started instead of starting a new one for the same question.

 

[quantumdude]

So I used the equation:
V = (k)(Q/r + Q/r)
V = (9.0E9)* [(2*1.6E-19)]/1.32e-15m + (2*1.6E-19)]/1.32E-15m)
V = 1454545.455 V

Note that this is an electric potential, not a potential energy.

So i need to convert to an eV so i divided by e which is 1.6E-19 as my professor said and other people on the forum and i got:
V = 9.0909E24 eV which is wrong! any ideas?

You don't convert from Volts to electron Volts. They have different dimensions. You convert from Joules to electron Volts. eV is a measure of energy.

So, go back to your calculations for potential energy (which had units of Joules) and convert those to eV.

 

[mr_coffee]

Sorry about the repost, i couldn't find my thread! I just saw Doc's responce, well if V = Kq/R, which is a V or a J/C, I don't see how I'm messing up, is it because I'm adding both pair and doc said i should only use 1, meaning V = KQ/r then i'll still end up with a J/C, why did my professor tell me to divide by e to convert from V to eV?

 

[Doc Al]

As Tom mentioned above, V is the potential, not the potential energy; to get the energy, you have to multiply by the charge, PE = Vq. Given the electric potential due to the first quark at the position of the second ($V = k q_1 /r$), find the potential energy by $k q_1 q_2 /r$. (That's for part (a))

 

[mr_coffee]

ohh i c, so I'm suppose to only use $V = k q_1 /r$ then multiply that by the charge itself, to get PE? Then do i still have to divide by e to conver to eV

 

[Doc Al]

To convert from standard units (J) to eV, you divide by the charge of the electron in Coulombs; in other words 1 eV = 1.60E-19 J. But that conversion should be very easy, since the quark charges are given in terms of electron charge.

 

[quantumdude]

Sorry about the repost, i couldn't find my thread!

If you go to your Profile you can click on "Find all threads started by mr_coffee", and you'll see them (with links) displayed on a page.

 

[mr_coffee]

maybe u have a differnet option because ur a mod, i don't see that option in user cp

 

[quantumdude]

My mistake, it's not the User CP. It's your Public Profile page. You can get there by clicking on your own username on any of your posts and selecting, "View Public Profile". You will then be directed to a page that let's you search for your own posts and threads. It's pretty handy.