NEET-XII-Physics
29: Electric Field and Potential
- #21Two charged particles with charge 2.0 × 10-8 C each are joined by an insulating string of length 1 m and the system is kept on a smooth horizontal table. Find the tension in the string.Ans : Given:
Magnitude of the charges, q = 2.0 × 10-8 C
Separation between the charges, r = 1 m
The tension in the string will be same as the electrostatic force between the charged particles.
So, `` T=F``
By Coulomb's Law,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}^{2}}{{r}^{2}}``
`` \Rightarrow T=\frac{9\times {10}^{9}\times {\left(2.0\times {10}^{-8}\right)}^{2}}{{1}^{2}}``
`` =3.6\times {10}^{-6}\,\mathrm{\,N\,}``
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