Test: Basic Properties of Charge & Coulomb’s Law (26 Oct) - JEE MCQ

The value of k = (1/4πε₀) = 8.854 x 10^-12 C² N^-1 m^-2 where ε₀ is permittivity of free space.

Charge of proton is q_P = 1.6 x 10^-19 C

Distance between the protons is, r = 3 x 10^-15m
The magnitude of electrostatic force between protons is 

The charge given out in one second = 1.6 x 10^-19 C x 10⁹ = 1.6 x 10^-10C
Time required to accumulate a charge of 1 C
= 1/(1.6 x 10^-10) = 6.25 x 10⁹ s = 198 years

By quantisation of charge, q = ne
or 

Electrostatic forces are both attractive and repulsive depending upon the type of charge, but gravitational forces is always attractive.

Here, q₁ = 1 x 10^-7 C, q₂ = 2 x 10^-7 C, r = 20cm = 20 x 10^-2 m

Total charge, q = 2.4 C
Then by quantization of charge, q = ne
∴ number of electrons, n = q/e = 2.4C/1.6 x 10^-19 C = 1.5 x 10¹⁹

 Here, q = -6 x 10^-7 C
∴ Number of electrons transferred from wool to polythene, n = 

Mass of water = 250 g,
Molecular mass of water = 18 g
Number of molecules in 18 g of water = 6.02 x 10²³
Number of molecules in one cup of water = (250/18) x 6.02 x 10²³
Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.
∴ Total positive charge present in one cup of water
= (250/18) x 6.02 x 10²³ x 10 x 1.6 x 10^-19 C = 1.34 x 10⁷C.