Three identical particles, each possessing the mass $m$ and charge $+q$, are placed at the corners of an…
Three identical particles, each possessing the mass $m$ and charge $+q$, are placed at the corners of an equilateral triangle with side $r_{0} .$ The particles are simultaneously set free and start flying apart symmetrically due to Coulomb's repulsion forces. The work performed by Coulomb's forces acting on each particle until the particles fly from one another to a very large distance is (where $\mathrm{k}=1 / 4 \pi \varepsilon_{0} .$ )
$\frac{3 k q^{2}}{r_{0}}$
$\frac{k q^{2}}{r_{0}}$
$\frac{3 k q^{2}}{2 r_{0}}$
$\frac{k q^{2}}{2 r_{0}}$
Solution
Since the given system is closed, the increase in $\mathrm{KE}$ is equal to decrease in P.E. $\Rightarrow \frac{3}{2} m v^{2}=\frac{2 k q^{2}}{r_{0}}-\frac{3 k q^{2}}{r}$
$\Rightarrow v=\sqrt{\frac{2 k q^{2}\left(r-r_{0}\right)}{m r r_{0}},}$
v will be max when $r \rightarrow \infty$ $\Rightarrow v_{\max }=\sqrt{\frac{2 k q^{2}}{m r_{0}}}$
The work performed by the interaction force
during the variation of the system's configuration is equal to the decrease in the potential energy $W=U_{1}-U_{2}=\frac{3 k q^{2}}{r_{0}}$
$\therefore$ Work done per particle $=\frac{k q^{2}}{r_{0}}$