NEET-XII-Physics
29: Electric Field and Potential
- #8Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?Ans : We know that minimum charge on a body (q) = charge on an electron
⇒ q = 1.6 × 10-19 C
Given:
Separation between the charges, r = 1 cm = 10-2 m
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
`` \Rightarrow F=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left({10}^{-2}\right)}^{2}}``
`` \Rightarrow F=2.3\times {10}^{-24}\,\mathrm{\,N\,}``
Page No 121: