Potential energy

Question

A charged particle q is shot towards another charged particle Q which is fixed, with a speed $\nu $. It approaches Q upto a closest distance r and then returns. If q were given a speed 2$\nu $ , the closest distances of approach would be

Moderate

Solution

Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. $\frac{1}{2}{\mathrm{mv}}^{2}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}.\frac{\mathrm{qQ}}{\mathrm{r}}$

Therefore the distance of closest approach is given by $\mathrm{r}=\frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon}}_{0}}.\frac{2}{{\mathrm{mv}}^{2}}\Rightarrow \mathrm{r}\propto \frac{1}{{\mathrm{v}}^{2}}$

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