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Surface Energy

Two electrons...

Question

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

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Solution

Let their velocities at 20cm a part
be v. Then
$Δ K E_{T o t a l} = 2 [\frac{M e}{2} v^{2} - \frac{m}{2} (0)^{2}] = M e v^{2}$
Potential energy 10 cm (U Initial )
$= \frac{- 1 k e^{2}}{\frac{10}{100}} = 10 k e^{2}$
Potential Energy 20 cm (U final )
$= \frac{k e^{2}}{20 / 100} = 5 k e^{2}$
$Δ U_{T o t a l} = U_{f i n a l} - U_{i n i t i a l}$
$= 5 k e^{2} - 10 k e^{2} = - 5 k e^{2}$
As electrostatics force is
conservative,
$Δ K E_{T o t a l} + Δ P E_{T o t a l} = 0$
$\Rightarrow m e v^{2} - 5 k e^{2} = 0$
$\Rightarrow v = \sqrt{\frac{5 k e^{2}}{m e}} = 35.58 m / s$

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