Two point charges q₁ and q₂ are fixed at a distance 3.0 cm as shown in the figure. A dust particle with mass $\mathrm{m}=5.0\times {10}^{-9} \mathrm{kg}$ and charge ${\mathrm{q}}_0=2.0\mathrm{nC}$ 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, $\mathrm{v}=\sqrt{\frac{2{\mathrm{q}}_0\left({\mathrm{V}}_{\mathrm{i}}-{\mathrm{V}}_{\mathrm{f}}\right)}{\mathrm{m}}}$            …(i) 

Electric potential at $\text{'}\mathrm{a}\text{'}:{\mathrm{V}}_{\mathrm{i}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_{1\mathrm{a}}}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_{2\mathrm{a}}}$

${\mathrm{V}}_{\mathrm{i}}=\left(9.0\times {10}^9\right)\times \left(\frac{3.0\times {10}^{-9}}{0.010}+\frac{\left(-3.0\times {10}^{-9}\right)}{0.020}\right)=1350 \mathrm{V}$      …(ii)

Electric potential at ${ }^{\text{'}}{\mathrm{b}}^{\mathrm{\prime }}: {\mathrm{V}}_{\mathrm{f}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_{1\mathrm{b}}}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_{2\mathrm{b}}}$

${\mathrm{V}}_{\mathrm{f}}=\left(9.0\times {10}^9\right)\times \left(\frac{3.0\times {10}^{-9}}{0.020}+\frac{\left(-3.0\times {10}^{-9}\right)}{0.010}\right)=-1350 \mathrm{V}$       …(iii)

From (i), (ii) and (iii) we get 

$\mathrm{v}=\sqrt{\frac{2\left(2.0\times {10}^{-9}\right)\left[\right(1350)-(-1350\left)\right]}{5.0\times {10}^{-9}}}=46.48 \mathrm{m}/\mathrm{s}$

rounding off to nearest integer