Question
At what distance should two charges, each equal to $1C,$ be placed so that the force between them equals your weight?
✓
Answer
Given: $q_1 = q_2 = 1C$ By Coulomb's law, the force of attraction between the two charges is given by
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$=\frac{9\times10^9\times1\times1}{\text{r}^2}$
However, the force of attraction is equal to the weight $(F = mg).$
$\therefore\text{mg}=\frac{9\times10^9}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{9\times10^9}{\text{m}\times10}=\frac{9\times10^8}{\text{m}} ($Taking $g = 10\ m/s^2)$
$\Rightarrow\text{r}^2=\frac{9\times10^8}{\text{m}}$
$\Rightarrow\text{r}=\frac{3\times10^4}{\sqrt{\text{m}}}$
Assuming that $m = 81\ kg,$
we have:$\text{r}=\frac{3\times10^4}{\sqrt{81}}$
$=\frac{3}{9}\times10^4\text{m}$
$=3333.3\text{m}$
$\therefore$ The distance $r$ is $3333.3m.$
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$=\frac{9\times10^9\times1\times1}{\text{r}^2}$
However, the force of attraction is equal to the weight $(F = mg).$
$\therefore\text{mg}=\frac{9\times10^9}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{9\times10^9}{\text{m}\times10}=\frac{9\times10^8}{\text{m}} ($Taking $g = 10\ m/s^2)$
$\Rightarrow\text{r}^2=\frac{9\times10^8}{\text{m}}$
$\Rightarrow\text{r}=\frac{3\times10^4}{\sqrt{\text{m}}}$
Assuming that $m = 81\ kg,$
we have:$\text{r}=\frac{3\times10^4}{\sqrt{81}}$
$=\frac{3}{9}\times10^4\text{m}$
$=3333.3\text{m}$
$\therefore$ The distance $r$ is $3333.3m.$
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