Energy Stored Charge Configuration w/ Picture

  • #1
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Homework Statement



If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to kq^2/L. In the space provided, enter the numeric value that multiplies the above factor, in W.

Homework Equations



Vba = Vb - Va = -Wba / q


The Attempt at a Solution



I need to sum the electric potentials to find the electric potential at one point. The potential due to charge A before "B" is placed there is:

kq / L

The potential due to charge B before "C" is placed is:

where I'm having problems. It states the point charges were initially infinitely far apart so I'm not sure how that affects "L."
 

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Answers and Replies

  • #2
SammyS
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Homework Statement



If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to kq^2/L. In the space provided, enter the numeric value that multiplies the above factor, in W.

Homework Equations



Vba = Vb - Va = -Wba / q

The Attempt at a Solution



I need to sum the electric potentials to find the electric potential at one point. The potential due to charge A before "B" is placed there is:

kq / L

The potential due to charge B before "C" is placed is:

where I'm having problems. It states the point charges were initially infinitely far apart so I'm not sure how that affects "L."
Do you know how to determine W?

If all the charges are initially infinitely far apart:
How much work does it take to bring charge A into position?

How much work does it take to bring charge B into position, assuming that charge A is already in position?

How much work does it take to bring charge C into position, assuming that charges A & B are already in position?

How much work does it take to bring charge D into position, assuming that charges A, B & C are already in position?​
 
  • #3
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Hi.

I think I know how to calculate work. For charge "C," I was thinking this:

(q) Vc - Va = -Wca


It took "A" 0 J of work because that's the origin. I thought it would take kq^2/L to bring charge "C" into place because that's what it took "B." Is this the incorrect way to think of this?
 
  • #4
SammyS
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Hi.

I think I know how to calculate work. For charge "C," I was thinking this:

(q) Vc - Va = -Wca


It took "A" 0 J of work because that's the origin. I thought it would take kq^2/L to bring charge "C" into place because that's what it took "B." Is this the incorrect way to think of this?
I'm assuming all charges have a magnitude of q.

It takes 0 J of work to bring charge A into position (if it's the first charge), because there is no other charge present at that time.

It then takes [itex]\displaystyle \frac{k\,q^2}{L}[/itex] Joules of work to bring charge B into position.

Once charge C is in position, how far is it from charge A?
How far from charge B?​
 
  • #5
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Once charge "C" is placed, it is now 1L away from "B" but "2L" away from "A."

∴ 0.5 * k^2q / L - 0 = 0.5 k^2q / L
 
  • #6
SammyS
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Once charge "C" is placed, it is now 1L away from "B" but "2L" away from "A."

∴ 0.5 * k^2q / L - 0 = 0.5 k^2q / L

No, C is [itex]\sqrt{2}\,L[/itex] away from A .

So how much work dies it take to move C there ?
 
  • #7
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I do not understand why "C" is 2^0.5 away from "A."

So W = Fd = Uc - Ua

So now I need to find the potential energy at "C" and the potential energy at "A" and find the difference.
 
  • #8
ehild
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I do not understand why "C" is 2^0.5 away from "A."

A and C are at the end points of the diagonal. If L is the length of one side of the square, what is the length of the diagonal?

So W = Fd = Uc - Ua

So now I need to find the potential energy at "C" and the potential energy at "A" and find the difference.

Uc-Ua is the work done if you move the charge from A to C. But you need to determine the work needed when a charge is brought in to point C from infinity. And that work is exactly the potential at C multiplied by the charge q.

ehild
 
  • #9
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Almost here.

To put "A" into position: 0 J
To put "B" into position: kq^2 / L
To put "C" into position: kq^2 / sqrt 2 * L
To put "D" into position: -kq^2 / L

0 J + kq^2 / L + kq^2 / sqrt 2 * L + (-kq^2 / L) =

kq^2 / L + kq^2 / sqrt 2 * L =

1 / 1 + 1 / sqrt 2 *(kq^2 / L) =

sqrt 2 / sqrt 2 + 1 / sqrt 2 =

1 + sqrt 2 / sqrt 2 (kq^2/L)

I know that was a pain to read but I'm still unsure of where I went wrong.
 
  • #10
ehild
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Almost here.

To put "A" into position: 0 J
To put "B" into position: kq^2 / L
To put "C" into position: kq^2 / sqrt 2 * L
To put "D" into position: -kq^2 / L

When you put C into position both A and B are already there, so both exert force on C.
The same with D: A, B, C are there. You need to find the contribution of all the three other charges to D-s potential energy.

ehild
 
  • #11
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I won't go through all of math but I found the answer to be: W = 0 J.
 
  • #12
ehild
Homework Helper
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I won't go through all of math but I found the answer to be: W = 0 J.

Correct! :smile:

ehild
 

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