CBSE-XI-Physics
04: The Forces
- #3At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?Ans : Given: `` {q}_{1}={q}_{2}=1\,\mathrm{\,C \,}``
By Coulomb's law, the force of attraction between the two charges is given by
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\in }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
`` =\frac{9\times {10}^{9}\times 1\times 1}{{r}^{2}}``
However, the force of attraction is equal to the weight (F = mg).
`` \therefore mg=\frac{9\times {10}^{9}}{{r}^{2}}``
`` \Rightarrow {r}^{2}=\frac{9\times {10}^{9}}{m\times 10}=\frac{9\times {10}^{8}}{m}(\,\mathrm{\,Taking \,}g=10\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2})``
`` \Rightarrow {r}^{2}=\frac{9\times {10}^{8}}{m}``
`` \Rightarrow r=\frac{3\times {10}^{4}}{\sqrt{m}}``
Assuming that m = 81 kg, we have:
`` r=\frac{3\times {10}^{4}}{\sqrt{81}}``
`` =\frac{3}{9}\times {10}^{4}\,\mathrm{\,m \,}``
`` =3333.3\,\mathrm{\,m \,}``
∴ The distance r is 3333.3 m.