Question

Two electrons lying $10cm$ apart are released. What will be their speed when they are $20cm$ apart?

Answer 1

Let their velocities at 20cm a part

be v. Then

$\Delta K{E}_{Total}=2[\frac{Me}{2}{v}^2-\frac{m}{2}(0{)}^2]=Me{v}^2$

Potential energy 10 cm (U Initial )

$=\frac{-1k{e}^2}{\frac{10}{100}}=10k{e}^2$

Potential Energy 20 cm (U final )

$=\frac{k{e}^2}{20/100}=5k{e}^2$

$\Delta U_{Total}=U_{final}-U_{initial}$

$=5k{e}^2-10k{e}^2=-5k{e}^2$

As electrostatics force is

conservative,

$\Delta K{E}_{Total}+\Delta P{E}_{Total}=0$

$\Rightarrow me{v}^2-5k{e}^2=0$

$\Rightarrow v=\sqrt{\frac{5k{e}^2}{me}}=35.58m/s$