Question
At what separation should two equal charges, 1.0C each, be placed so that the force between them equals the weight of a 50kg person?
✓
Answer
Given: Magnitude of charges, $\text{q}_1=\text{q}_2=1\text{C}$ Electrostatic force between them, F = Weight of a 50kg person$\text{mg}=50\times9.8=490\text{N}$
$\text{mg}=490\text{N}$
Let the required distance be r. By Coulomb's Law, electrostatic force,$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow490=\frac{9\times10^9\times1\times1}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{9\times10^9}{490}$
$\Rightarrow\text{r}=\sqrt{\frac{9}{49}\times10^8}$
$=\frac{3}{7}\times10^4\text{m}$
$=4.3\times10^3\text{m}$
$\text{mg}=490\text{N}$
Let the required distance be r. By Coulomb's Law, electrostatic force,$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow490=\frac{9\times10^9\times1\times1}{\text{r}^2}$
$\Rightarrow\text{r}^2=\frac{9\times10^9}{490}$
$\Rightarrow\text{r}=\sqrt{\frac{9}{49}\times10^8}$
$=\frac{3}{7}\times10^4\text{m}$
$=4.3\times10^3\text{m}$
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