Question

Two particles of masses 5.0 g each and opposite charges of +4.0 × 10⁻⁵ C and −4.0 × 10⁻⁵ C are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm.

Answer 1

Given:
Magnitude of charges, q = 4.0 × 10⁻⁵ C
Initial separation between charges, r = 1 m
Initial speed = 0; so, initial K.E. = 0
Mass of the particles, m = 5.0 g =0.005 kg
Let the required velocity of each particle be v.
By the law of conservation of energy,
Initial P.E. + Initial K.E. = Final P.E. + Final K.E.
$$\begin{gathered} \frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r}=2\times \frac{1}{2}m{v}^2+\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r/2} \\ \Rightarrow \frac{-1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r}=m{v}^2-\frac{2}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r} \\ \Rightarrow m{v}^2=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q^2}{r} \\ \Rightarrow v=\sqrt{\frac{1}{4\mathrm{\pi }{\epsilon }_{0}m}\frac{q^2}{r}} \\ \Rightarrow v=\sqrt{\frac{9\times {10}^{-9}\times {\left(4\times {10}^{-5}\right)}^2}{0.005\times 1}} \\ \Rightarrow v=53.66\mathrm{m}/\mathrm{s} \end{gathered}$$