The Forces — Physics STD 11 Science — Question

$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$

$=\frac{9\times10^9\times1\times1}{\text{r}^2}$

$\therefore\text{mg}=\frac{9\times10^9}{\text{r}^2}$

$\Rightarrow\text{r}^2=\frac{9\times10^9}{\text{m}\times10}=\frac{9\times10^8}{\text{m}}$ (Taking g = 10m/s²)

$\Rightarrow\text{r}^2=\frac{9\times10^8}{\text{m}}$

$\Rightarrow\text{r}=\frac{3\times10^4}{\sqrt{\text{m}}}$

$\text{r}=\frac{3\times10^4}{\sqrt{81}}$

$=\frac{3}{9}\times10^4\text{m}$

$=3333.3\text{m}$

$\therefore$ The distance r is 3333.3m.