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

Question

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Answer 1

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 )
$= \frac{- 1 k e ^{2}}{\frac{1 0}{1 0 0}} = 10 k e^{2}$
Potential Energy 20 cm (U final )
$= \frac{k e ^{2}}{2 0 / 1 0 0} = 5 k e^{2}$
$Δ U_{T o t a l} = U_{f i n a l} - U_{i n i t i a l}$
$= 5 k e^{2} - 10 k e^{2} = - 5 k e^{2}$
As electrostatics force is
conservative,
$Δ K E_{T o t a l} + Δ P E_{T o t a l} = 0$
$\Rightarrow m e v^{2} - 5 k e^{2} = 0$
$\Rightarrow v = \frac{5 k e ^{2}}{m e} = 35.58 m / s$

Answer 2

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{1 0}{1 0 0}} = 10 k e^{2}

Potential Energy 20 cm (U final )

= \frac{k e ^{2}}{2 0 / 1 0 0} = 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 = \frac{5 k e ^{2}}{m e} = 35.58 m / s