Electron repulsion work problem.

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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