Question

Two identical particles, each with a charge of 2.0 × 10⁻⁴ C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speed of the particles when the separation becomes large?

Answer 1

Given:
Magnitude of charges, q = 2.0 × 10⁻⁴ C
Mass of particles, m = 10 g = 0.01 kg
Separation between the charges, r = 10 cm = 0.1 m
Force of repulsion,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r^2}$
$$\begin{gathered} F=1.8\times {10}^4\mathrm{N} \\ F=m\times a \\ \Rightarrow a=1.8\times {10}^6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Now,
$$\begin{gathered} v^2=u^2+2as \\ {v}^2=0+2\times 1.8\times {10}^6\times 10\times {10}^{-2} \\ v=\sqrt{36\times {10}^4} \\ v=600\mathrm{m}/\mathrm{s} \end{gathered}$$