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## Homework Statement

A point charge Q is fixed in position, and a second object with charge q and mass m moves directly toward it from a great distance. If the initial speed of the object is v, compute the minimum distance between the two objects. If Q=1.0 μC, q=1.0 nC, m=1.0*10

^{5}kg, and v= 3.0*10

^{5}m/s, what is the minimum distance of approach?

## Homework Equations

## The Attempt at a Solution

[tex]\begin{array}{l}

\frac{{kQq}}{{r_1 }} + \frac{1}{2}mv_0^2 = \frac{{kQq}}{{r_2 }} + \frac{1}{2}v_f^2 ,\,\,\,\,\,\,\,\,\,\,\,\,v_f = 0 \\

\\

\frac{{kQq}}{{r_1 }} + \frac{1}{2}mv_0^2 = \frac{{kQq}}{{r_2 }} \\

\\

r_2 = \frac{{kQq}}{{\left( {\frac{{kQq}}{{r_1 }} + \frac{1}{2}mv_0^2 } \right)}} \\

\\

r_2 = \frac{{8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }}1.0\mu {\rm{C}} \cdot 1.0{\rm{nC}}}}{{\left( {\frac{{8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }}1.0\mu {\rm{C}} \cdot 1.0{\rm{nC}}}}{{r_1 }} + \frac{1}{2}1.0 \times 10^{ - 5} \cdot 3.0 \times 10^5 } \right)}} \\

\end{array}[/tex]

This reminds me somewhat of gravity's escape velocity problem, where your final distance is irrelevant as long as it is large (theoretically infinity).

How do I dismiss r1, the initial starting position in this problem? Can I just get rid of the entire first term in the denominator, since when r1 is large, it approaches 0?

Thanks!

**edit, 3*10

^{5}should be (3*10

^{5})

^{2}. I get 2*10

^{-11}meters for the answer. Did I do this right?

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