A particle of mass m carrying charge 'q' is projected with velocity ‘v’ from point 'A' towards an infinite line of charge from a distance ‘a’. Its speed reduces to zero momentarily at point 'B' which is at a distance $$a/2$$ from the line of charge. If another particle with mass m and charge '$$-q$$' is projected with the same velocity 'v' from 'A' towards the line of charge. If $$v=102$$ $$m/s$$, what will be its speed (in$$ $$m/s) at 'B' ?

Question

A particle of mass m carrying charge 'q' is projected with velocity ‘v’ from point 'A' towards an infinite line of charge from a distance ‘a’. Its speed reduces to zero momentarily at point 'B' which is at a distance $$a/2$$ from the line of charge. If another particle with mass m and charge '$$-q$$' is projected with the same velocity 'v' from 'A' towards the line of charge. If $$v=102$$ $$m/s$$, what will be its speed (in$$ $$m/s) at 'B' ?

Infinity Learn · Accepted Answer

Applying conservation of mechanical energy

$ΔK + ΔU = 0 or (K_{f} - K_{i}) + q (V_{f} - V_{i}) = 0$ …(i)

When a positive charge is projected towards the wire

$(0 - \frac{1}{2} {mv}^{2}) + qΔV = 0 \Rightarrow \frac{1}{2} {mv}^{2} = q (V_{B} - V_{A})$ …(ii)

When a negative charge is projected towards the wire

$(\frac{1}{2} {mv}_{B}^{2} - \frac{1}{2} {mv}^{2}) + (- q) ΔV = 0$

$\Rightarrow (\frac{1}{2} {mv}_{B}^{2} - \frac{1}{2} {mv}^{2}) = q (V_{B} - V_{A})$ …(iii)

From (ii) and (iii), we get

$\frac{1}{2} {mv}^{2} = (\frac{1}{2} {mv}_{B}^{2} - \frac{1}{2} {mv}^{2}) \Rightarrow v_{B}^{2} = 2 v^{2} or v_{B} = \sqrt{2} v$

$= \sqrt{2} \times 10 \sqrt{2} = 20 m / s$

Electrostatic potential energy

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