Two-point charges q₁ and q₂ are fixed at a distance 3.0 cm as shown in the figure. A dust particle with mass m = 5.0 x 10^-9 kg and charge q₀ = 2.0 nC starts from rest at point 'a' and moves in a straight line to point ' b'.

What is its speed v at point b (in m/s) ?

Applying conservation of mechanical energy$\mathrm{\Delta K}+\mathrm{\Delta U}=0$
or $\left({\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\right)+\left({\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}\right)=0$Here, ${\mathrm{K}}_{\mathrm{i}}=0\text{ and }{\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}^2\text{ and }{\mathrm{U}}_{\mathrm{i}}={\mathrm{q}}_0{\mathrm{V}}_{\mathrm{i}}\text{ and }{\mathrm{U}}_{\mathrm{f}}={\mathrm{q}}_0{\mathrm{V}}_{\mathrm{f}}$
Hence, $\left({\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\right)+{\mathrm{q}}_0\left({\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}}\right)=0\left(\frac{1}{2}{\mathrm{mv}}^2-0\right)+\left({\mathrm{q}}_0{\mathrm{V}}_{\mathrm{f}}-{\mathrm{q}}_0{\mathrm{V}}_{\mathrm{i}}\right)=0$
And solving for v, we find,$\mathrm{v}=\sqrt{\frac{2{\mathrm{q}}_0\left({\mathrm{V}}_{\mathrm{i}}-{\mathrm{V}}_{\mathrm{f}}\right)}{\mathrm{m}}}$.........(1)
Electric potential at ‘a’${\mathrm{V}}_{\mathrm{i}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_{1\mathrm{a}}}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_{2\mathrm{a}}}$
${\mathrm{V}}_{\mathrm{i}}=\left(9.0\times {10}^9\right)\times \left(\frac{3.0\times {10}^{-9}}{0.010}+\frac{\left(-3.0\times {10}^{-9}\right)}{0.020}\right)=1350\mathrm{V}\dots .\left(\mathrm{ii}\right)$
$V_f=\left(9.0\times {10}^9\right)\times \left(\frac{3.0\times {10}^{-9}}{0.020}+\frac{\left(-3.0\times {10}^{-9}\right)}{0.010}\right)=-1350\mathrm{V}.....\left(\mathrm{iii}\right)$
From (i) (ii) and (iii) we get$\mathrm{v}=\sqrt{\frac{2\left(2.0\times {10}^{-9}\right)\left[\right(1350)-(-1350\left)\right]}{5.0\times {10}^{-9}}}=46\mathrm{m}/\mathrm{s}$