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A particle of mass $100\, gm$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force........$m/s$
$-1.73$
$2.73$
$-2.73$
$1.73$
Solution
Loss of potential energy $=$ gain in kinetic energy
$\mathrm{U}_{1}-\mathrm{U}_{2}=\Delta \mathrm{K}.$
$\mathrm{KQq}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]=1 / 2 \mathrm{mv}^{2}-0$
$\mathrm{v}=\sqrt{\frac{2 \mathrm{k} \mathrm{Q} \mathrm{q}}{\mathrm{m}}\left[\frac{\mathrm{r}_{2}-\mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right]}$
$=\sqrt{\frac{2 \times 9 \times 10^{9} \times 5 \times 10^{-6} \times 2 \times 10^{-6} \times 2.5}{0.1 \times 3 \times 0.5}}=1.73 \mathrm{\,m} / \mathrm{s}$
