NEET-XII-Physics
29: Electric Field and Potential
- #12Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?Ans : Given:
Force of attraction between the spheres, F = 0.1 N
Separation between the spheres, r = 10-2 m
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
Let the no. of electrons transferred from one sphere to the other be n. Then,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{\left(nq\right)}^{2}}{{r}^{2}}``
`` \Rightarrow 0.1=\frac{9\times {10}^{-9}\times {n}^{2}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{10}^{-4}}``
`` \Rightarrow {n}^{2}=\frac{0.1\times {10}^{-4}}{9\times {10}^{9}\times 1.6\times 1.6\times {10}^{-38}}``
`` \Rightarrow n=2\times {10}^{11}``
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