Electrical Potential

  • Thread starter stylez03
  • Start date
  • #1
139
0

Homework Statement


Three equal point charges, each with a charge of 2.00 uC, are placed at the corners of an equilateral triangle whose sides have a length of 0.400 m. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)


Homework Equations



[tex] U_{tot} = \frac {Q^{2}} {4*pi*8.85*10^{-12}} * (\frac {A} {d1} + \frac {B} {d2} + \frac {C} {d3}) [/tex]


The Attempt at a Solution



[tex] U_{tot} = \frac {(2.00*10)^{-19}} {4*pi*8.85*10^{-12}} * (\frac {4} {0.400} + \frac {4} {0.400} + \frac {4} {0.400}) [/tex]

The online program says i'm off by an additive constant once again and my work seems right?
 

Answers and Replies

  • #2
340
0
Think about assembling the charges one by one, bringing them each in from infinitely far away.
 
  • #3
139
0
Think about assembling the charges one by one, bringing them each in from infinitely far away.
When my professor tells us to think about things, usually it involves something going to 0? Are the charges 0 as you bring them closer?
 
  • #4
340
0
No. How much work does it take to bring in one charge from infinity if there are no other charges around?
 
  • #5
139
0
No. How much work does it take to bring in one charge from infinity if there are no other charges around?
The work done would be U_a - U_b = -(U_b-U_a) = -ChangeU
 
  • #6
340
0
If there are no other charges around, there is no potential which you would have to do work against to bring in the first charge. Now what about the second charge?
 
  • #7
139
0
If there are no other charges around, there is no potential which you would have to do work against to bring in the first charge. Now what about the second charge?
The work done should be the amount of work it takes to bring in the second charge?
 
  • #8
340
0
Calculate that work. Now you have two charges in a region. How much energy does it take to bring in the third charge? Then add.
 
  • #9
139
0
Calculate that work. Now you have two charges in a region. How much energy does it take to bring in the third charge? Then add.
Is there an equation for work in this case? I thought my original equation was it?
 
  • #10
139
0
Should it not be:

[tex] U_{tot} = 3*( (\frac {1} {4*pi*8.85*10^{12}}) * (\frac {2.00*10^-{19}} {0.400}) )[/tex]


I know the amount of work is just the summation of amount of work to move one charge.
 
  • #11
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
Should it not be:

[tex] U_{tot} = 3*( (\frac {1} {4*pi*8.85*10^{12}}) * (\frac {2.00*10^-{19}} {0.400}) )[/tex]


I know the amount of work is just the summation of amount of work to move one charge.
No. As Robb suggested, think about assembling the array of charges. So, lets start with just one charge in free space; there is no external electric field and no other charges, so the work done to place the charge is zero. No consider bringing a second charge from d = [itex]\infty[/itex] to d = 0.4 How much work must you do on this second charge?
 
  • #12
139
0
No. As Robb suggested, think about assembling the array of charges. So, lets start with just one charge in free space; there is no external electric field and no other charges, so the work done to place the charge is zero. No consider bringing a second charge from d = [itex]\infty[/itex] to d = 0.4 How much work must you do on this second charge?
Amount of work would be:

[tex] V = \frac {K*Q} {r} [/tex]
 
  • #13
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
Amount of work would be:

[tex] V = \frac {K*Q} {r} [/tex]
Almost, but not quite. Note that electrical potential is not the same as electric potential energy.
 
  • #14
139
0
Almost, but not quite. Note that electrical potential is not the same as electric potential energy.
[tex] U = \frac {1} {4*pi*E_{0}} * \frac {Q_{1}*Q_{2}} {r} [/tex]
 
  • #15
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
[tex] U = \frac {1} {4*pi*E_{0}} * \frac {Q_{1}*Q_{2}} {r} [/tex]
Correct, and since Q1 = Q1 := Q, we can say that

[tex]U = k\frac{Q^2}{r}[/tex]

So, now we have two charges in place and have done U joules of work. So how much work must we do on the third charge (Q) to bring it from d = [itex]\infty[/itex] to d = 0.4?
 
  • #16
139
0
Correct, and since Q1 = Q1 := Q, we can say that

[tex]U = k\frac{Q^2}{r}[/tex]

So, now we have two charges in place and have done U joules of work. So how much work must we do on the third charge (Q) to bring it from d = [itex]\infty[/itex] to d = 0.4?
[tex]U = k\frac{Q^3}{r}[/tex]
 
  • #17
340
0
Careful, your units aren't even correct here.
First, conceptually, will it take more work to bring in the third charge compared to the second, or less work?
 
  • #18
139
0
Careful, your units aren't even correct here.
First, conceptually, will it take more work to bring in the third charge compared to the second, or less work?
You just said this equation was right, now you say it's incorrect.
 
  • #19
340
0
Umm, Q*Q is not the same units as Q*Q*Q.
 
  • #20
139
0
Umm, Q*Q is not the same units as Q*Q*Q.
Well I'm lost in what you're saying now.

You said to move 1 charge the potential is:

[tex]U = k\frac{Q^2}{r}[/tex]

After that I have no idea what you're talking about anymore
 
  • #21
340
0
[tex]Q^2[/tex]is not equal to [tex]Q^3[/tex]
 
Last edited:
  • #22
139
0
[tex]Q^2[/tex]is not equal to [tex]q^3[/tex]

sorry for some reason the latex is not editing correctly.
Okay fine, I understand by moving in another charge that it isn't multiplying by another charge, but I'm still not sure how much work it takes to add another? I thought you would just add the work of another charge as we found in the first charge, so it would be 2*U.
 
  • #23
340
0
After the first two are in place, the third charge has to be brought into a region of space where there are other charges that produce a potential. In essence, we must move that charge up the potential hill, just as you had to for the second charge.
The first charge we get for "free" since there is no potential (hill) to push the charge up. The second charge takes some work, since the first charge produces a potential (hill). How much- > [tex] V= kQ/r [/tex], which is the potential produced by the first charge so the work is [tex] U = qV [/tex] where q is the charge you are bringing up the potential V. Now the third charge must be brought up a potential produced by the first two charges. So find the new potential, [tex] V_{new} = V1 + V2[/tex]. Then the work for the thrid charge is [tex]U = q Vnew[/tex] where Vnew is the new potential created by the first two charges.
I hope that makes some sense, it is friday and my brain is fried!:surprised
 
Last edited:
  • #24
139
0
After the first two are in place, the third charge has to be brought into a region of space where there are other charges that produce a potential. In essence, we must move that charge up the potential hill, just as you had to for the second charge.
The first charge we get for "free" since there is no potential (hill) to push the charge up. The second charge takes some work, since the first charge produces a potential (hill). How much- > [tex] V= kQ/r [/tex], which is the potential produced by the first charge so the work is [tex] U = qV [/tex] where q is the charge you are bringing up the potential V. Now the third charge must be brought up a potential produced by the first two charges. So find the new potential, [tex] V = V1 + V2[/tex]. Then the work for the thrid charge is [tex]U = q Vnew[/tex] where Vnew is the new potential created by the first two charges.
I hopw that makes some sense, it is friday and my brain is fried!:surprised
Yes it is friday, and my brain is fried as well. So let me get this right:

First charge required 0 amounts of work because there was no potential "hill" to push up. So the potential energy of the system is currently Zero

Second charge requires work because there is a potential "hill" to be pushed up, since charge 1 has a potential to produce.
Charge 1 produces [tex] V= kQ/r [/tex]

So the amount of work for Charge 2 => [tex] U = qV [/tex]

Third charge requires work too because now there is a potential "hill" from both Charge1 + Charge2 which is [tex] U = qVnew [/tex]

So is [tex] Vnew => V= kQ/r [/tex]

and

[tex] U = q*(kQ/r) [/tex]
 
  • #25
340
0
Almost correct, except that Vnew is due to both charges, so you add up the potential from each. V1+V2
 

Related Threads on Electrical Potential

Replies
4
Views
1K
Replies
8
Views
2K
Replies
4
Views
1K
Replies
1
Views
755
Replies
2
Views
2K
Replies
1
Views
1K
Replies
3
Views
6K
Replies
2
Views
12K
Replies
1
Views
7K
Top