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 $v = 10 \sqrt{2} m / s$ , what will be its speed (in m/s) at 'B' ?

Difficult

Solution

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
$\frac{1}{2} {mv}^{2} = (\frac{1}{2} {mv}_{B}^{2} - \frac{1}{2} {mv}^{2}) \Rightarrow v_{B}^{2} = 2 v^{2} or v_{B} = \sqrt{2} v$
$= \sqrt{2} \times 10 \sqrt{2} = 20 m / s$

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