Electron repulsion work problem.

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Homework Help Overview

The problem involves calculating the work done in moving a third electron from the origin to a point at (3,0) while two other electrons are fixed at (5,0) and (-5,0). The subject area pertains to electrostatics and the forces between charged particles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the distances involved and the forces acting on the third electron. There is an attempt to integrate the forces to find the work done. Some participants question the correctness of the calculations and the book's answer.

Discussion Status

The discussion has seen participants agreeing on the initial calculations, but later, one participant points out a potential error in the integration approach regarding the direction of forces. This has led to a reconsideration of the solution, with acknowledgment of the book's answer as possibly correct.

Contextual Notes

There is a noted concern about the forces not pointing in the same direction, which affects the integration process. Participants are also self-teaching and engaging with the material actively.

B3NR4Y
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The problem: If two electrons are held at fixed points (5,0) and (-5,0), respectively, find the work done in moving a third electron from the origin to (3,0)

All equations:
Force of repulsion between two electrons:
\begin{equation}
\frac{1}{d^{2}}.
\end{equation}
\begin{equation}
W = \int_{a}^{b} \, f(x) dx
\end{equation}
"D" is the distance between the electrons

Attempt at a solution:
The distance between the electron at (-5,0) and the third electron is, 5+x where x is the position of the electron at any point on its journey to (3,0). The distance between the electron at (5,0) and the third electron is (5-x). I found the work by adding the two work equations like this:
\begin{equation}
\begin{split}
W&= \int_{0}^{3} \frac{1}{(5+x)^{2}} dx + \int_{0}^{3} \frac{1}{(5-x)^{2}} dx \\
& = -[-\frac{1}{5-x} |^{3}_{0}] + [-\frac{1}{5+x} |^{3}_{0}\\
&=\frac{3k}{8} ergs
\end{split}
\end{equation}
Book says the answer is
\begin{equation}
\frac{9k}{40}\, ergs
\end{equation}
 
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B3NR4Y said:
The problem: If two electrons are held at fixed points (5,0) and (-5,0), respectively, find the work done in moving a third electron from the origin to (3,0)

All equations:
Force of repulsion between two electrons:
\begin{equation}
\frac{1}{d^{2}}.
\end{equation}
\begin{equation}
W = \int_{a}^{b} \, f(x) dx
\end{equation}
"D" is the distance between the electrons

Attempt at a solution:
The distance between the electron at (-5,0) and the third electron is, 5+x where x is the position of the electron at any point on its journey to (3,0). The distance between the electron at (5,0) and the third electron is (5-x). I found the work by adding the two work equations like this:
\begin{equation}
\begin{split}
W&= \int_{0}^{3} \frac{1}{(5+x)^{2}} dx + \int_{0}^{3} \frac{1}{(5-x)^{2}} dx \\
& = -[-\frac{1}{5-x} |^{3}_{0}] + [-\frac{1}{5+x} |^{3}_{0}\\
&=\frac{3k}{8} ergs
\end{split}
\end{equation}
Book says the answer is
\begin{equation}
\frac{9k}{40}\, ergs
\end{equation}

I agree with your answer.
 
Dick said:
I agree with your answer.

So probably just an error in the book?

Good :3
 
B3NR4Y said:
So probably just an error in the book?

Good :3

Yeah, I think so.
 
B3NR4Y said:
So probably just an error in the book?

Good :3

Hope it's not to late to change my mind. There is a problem. The forces you are integrating don't point in the same direction. You need a minus sign on one of your forces. The book answer is correct. Sorry about that!
 
Dick said:
Hope it's not to late to change my mind. There is a problem. The forces you are integrating don't point in the same direction. You need a minus sign on one of your forces. The book answer is correct. Sorry about that!

No worries! Thank you. Makes sense, the answer is correct now. I feel honored you answered me.
 
B3NR4Y said:
No worries! Thank you. Makes sense, the answer is correct now. I feel honored you answered me.

Glad you feel honored by a not very accurate response :). Hope that won't deter you from posting again.
 
Dick said:
Glad you feel honored by a not very accurate response :). Hope that won't deter you from posting again.

It most definitely will not. I'm 15 and self-teaching, so I will post here often. Thank you again.
 

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