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An electron is projected, with an initial speed of $1.10 \times 10^5 \text{m/s}$, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value?

i know it can be solved by equating the total energy as K.E+P.E(electron)=K.E+P.E(Proton) kinetic energy is $ \frac{1}{2}mv^2$. how can i find out potential energy of each particle?

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up vote 2 down vote accepted

The potential energy due to the electrostatic interaction between two particles of charges $q_1$ and $q_2$ is $$ U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} $$ where $r$ is the distance between them.

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Thank you..@joshphysics – sky rain Mar 6 '13 at 23:03
You're welcome @skyrain. – joshphysics Mar 6 '13 at 23:03

You have written down the right equation. Use the electric potential energy between the electron and the proton and write:

$\frac{1}{2}mv_1^2-\frac{e^2}{4\pi\epsilon_0}\frac{1}{R} =\frac{1}{2}mv_2^2-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}$

Put $v_2=2v_1$, ignore the $1/R$ term since $R$ is very large, so that the electron is in the classical energy region (i.e. in the continuous energy spectrum), and solve the resulting equation for $r$. You should find that the distance $r$ for which the speed of the electron doubles is inversely proportional to the initial speed. Substitute the physical constants into the equation you found to calculate the value of $r$. I think it should come out to be $\sim 1.4\times 10^{-8}$m.

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