NEET-XII-Physics
29: Electric Field and Potential
- #4Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges, so that the force between them equals the weight of a 50 kg person?Ans : Let the magnitude of each charge be q.
Separation between them, r = 1 m
Force between them, F = 50 × 9.8 = 490 N
By Coulomb's Law force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
`` \Rightarrow 490=9\times {10}^{9}\times \frac{{q}^{2}}{{1}^{2}}``
`` \Rightarrow {q}^{2}=54.4\times {10}^{-9}``
`` \Rightarrow q=\sqrt{54.4\times {10}^{-9}}=23.323\times {10}^{-5}\,\mathrm{\,C\,}``
`` \,\mathrm{\,Or\,}q=2.3\times {10}^{-4}\,\mathrm{\,C\,}``
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