Two charged particles each having charge ' $q$ ' and mass ' $m$ ' are held at rest while their separation is…

Two charged particles each having charge ' $q$ ' and mass ' $m$ ' are held at rest while their separation is ' $r$ '. The speed of the particles when their separation is ' $\frac{\mathrm{r}}{2}$, will be ( $\varepsilon_0=$ permittivity of the medium)
  1. $\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{mr}}$
  2. $\frac{\mathrm{q}}{2 \pi \varepsilon_0 \mathrm{mr}}$
  3. $\frac{\mathrm{q}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}$
  4. $\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \mathrm{mr}}$

Solution

By conservation of energy for the system of two charges, $(\text { K.E. + P.E. })_{\text {initial }}=(\text { K.E. }+ \text { P.E. })_{\text {final }}$
Initially the charges are at rest, $\mathrm{KE}_{\text {initial }}=0$, Final K.E. $=-\frac{1}{2}(2 \mathrm{~m}) \mathrm{v}^2$ ...(Like charges repel each other) $\begin{array}{ll} & \left(0+\left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}\right)\right)=\left(-\frac{1}{2}(2 m) v^2+\left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r / 2}\right)\right) \\ & \left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r}\right)=\left(-m v^2+\left(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r / 2}\right)\right) \\ & m v^2=\frac{q^2}{4 \pi \varepsilon_0}\left(\frac{2}{r}-\frac{1}{r}\right) \\ \therefore \quad & v=\frac{q}{\sqrt{4 \pi \varepsilon m r}} \end{array}$

Asked in: MHT CET 2024 (03 May Shift 2)

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