- #1

rpthomps

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- 19

## Homework Statement

Charge q1 is fixed to a spring with constant k_s. It is at equilibrium initially and located r+x away from q2 (which is oppositely charged). The charge q1 is released and allowed to move closer to q2, pulling the spring and stopping. Now the charges are only r distance away. I would like to assume I can solve this using both force and energy approaches. Here is my attempt...[/B]

## Homework Equations

## The Attempt at a Solution

Using forces...

[tex]

|F_{ e }|=|F_{ s }|\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ r^{ 2 } } =k_{ s }x\quad (1)

[/tex]

Using Energy

[tex]

\\ \\ r^{ ' }=r+\Delta x\\ E_{ T }=E_{ e }+E_{ s }\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ r^{ ' } } =\frac { k_{ e }q_{ 1 }q_{ 2 } }{ r } +\frac { 1 }{ 2 } k_{ s }\Delta x^{ 2 }\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ \Delta x } (\frac { 1 }{ r^{ ' } } -\frac { 1 }{ r } )=\frac { 1 }{ 2 } k_{ s }\Delta x\\ \frac { k_{ e }q_{ 1 }q_{ 2 } }{ \Delta x } (\frac { r-r^{ ' } }{ r^{ ' }r } )=\frac { 1 }{ 2 } k_{ s }\Delta x\\ k_{ e }q_{ 1 }q_{ 2 }(\frac { 2 }{ (r+\Delta x)r } )=k_{ s }\Delta x\quad (2)\\

[/tex]

To me, equation (1) and equation (2) should be the same. They are not. The only way they can be is if r=x. Any help would be greatly appreciated...

Ryan

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