The number of electric lines of force that radiate outwards from one coulomb of charge in vacuum is
- $1.13 \times 10^{11}$
- $1.13 \times 10^{10}$
- $0.61 \times 10^{11}$
- $0.61 \times 10^{9}$
Solution
Number of lines of force $=$ Electric force $=\frac{q}{\varepsilon_{0}}=\frac{1}{8.85 \times 10^{-12}}=1.13 \times 10^{11}$ *
Asked in: JEE Mains - Electrostatics - Test 2