Test: Coulomb's Law (NCERT) - NEET MCQ

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

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

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 

Here, q₁ = q₂ = 3.2 x 10^-7 C, r = 60cm = 0.6 m
Electrostatic force,

The force between two charges q1 and q2 at a distance r is given by Coulomb's law:

F = k*q1*q2/r^2

where k is Coulomb's constant (9*10^9 N*m^2/C^2).

Given that F = 10 N, q1 = 3 μC = 3*10^-6 C, and q2 = 4 μC = 4*10^-6 C, we can solve for r^2:

10 = 9*10^9 * 3*10^-6 * 4*10^-6 / r^2
r^2 = 9*10^9 * 3*10^-6 * 4*10^-6 / 10
r^2 = 1.08*10^-2 m^2

Now, if each charge is given an additional -6 μC, the new charges are q1' = -3 μC = -3*10^-6 C and q2' = -2 μC = -2*10^-6 C. The new force F' is:

F' = k*q1'*q2'/r^2
F' = 9*10^9 * -3*10^-6 * -2*10^-6 / 1.08*10^-2
F' = 5 N

So, the new force is 5N. Hence, option

Here, for an electron and a proton 

Here, q₁ = 0.2 μC = 0.2 x 10^-6 C
q₂ = 0.4 μC = -0.4 x 10^-6 C, F = -0.4 N
As F = 

= 1.8 x 10^-3
∴ r = (1.8 x 10^-3)^1/2 = 0.0424 m = 42.4 x 10^-3 m

The acceleration due to given coulombic force F is 
a = (F/m) or a ∝ (1/m) ...(1)
∴  where m_c and m_p are masses of electron and proton respectively