Question

Two 1 Coulomb charges are kept at 1m distance in air medium. Force of attraction or repulsion between them will be ________

(a) 9*10^9 N

(b) 1 dyne

(c) 1 N

(d) 3*10^3 N

The question was asked by my college professor while I was bunking the class.

My doubt stems from Coulomb’s Law in section Charges and Fields of Physics – Class 12

Answer 1

Right option is (a) 9*10^9 N

The best explanation: According to Coulomb’s Law, F=\(\frac {1}{4\pi \varepsilon_o }*\frac {q_1q_2}{r^2}\). And \(\frac {1}{4\pi \varepsilon_o}\) = 9*10^9N in the SI system. In this case, q1=q2=1C and r=1m. Now substituting the values we get F=9*10^9N. Similarly, the electric field at a distance of 1m from a 1C charge is 9*10^9N/C.