Question

a small ball of mass 50mg has a charge of +5 esu and velocity 10 cm/s.how close dose tha ball approach a charge of 10 esu ?

Answer 1

Suppose the distance between ball and charge is 'r'.
Using,
$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{4\mathrm{\pi }{\in }_0}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ \frac{1}{2}\times 0.05\times {10}^2=9\times {10}^9\times \frac{5\times 10}{{\mathrm{r}}^2} \\ \mathrm{or}{\mathrm{r}}^2=\frac{2\times 9\times {10}^9\times 5\times 10}{0.05\times {10}^2} \\ =\frac{90\times {10}^8}{0.05}=1800\times {10}^8 \\ \mathrm{or}\mathrm{r}=4.24\times {10}^5\mathrm{cm} \end{gathered}$$