Calculation of force between two point charges in vacuum

Given:
Distance between the two point charges in vacuum, r = 1mm = 0.001m
Force between the two point charges in vacuum, F = 18N

The force between two point charges in vacuum is given by Coulomb's law as:
F = k*q1*q2/r^2
where k is the Coulomb's constant, q1 and q2 are the magnitudes of the two point charges.

Substituting the given values, we get:
18 = k*q1*q2/(0.001)^2
or q1*q2 = 18*(0.001)^2/k

Calculation of force between two point charges with a glass plate

Now, a glass plate of thickness 1mm and dielectric constant 6 is kept between the two point charges. The force between the charges will change due to the presence of the glass plate.

The force between two point charges in a medium with dielectric constant is given by:
F = (1/4*pi*epsilon)*q1*q2/r^2
where epsilon is the permittivity of the medium, which is given by:
epsilon = epsilon0*epsilonr
where epsilon0 is the permittivity of vacuum and epsilonr is the relative permittivity or dielectric constant of the medium.

Substituting the given values, we get:
F = (1/4*pi*epsilon0*epsilonr)*q1*q2/r^2

For glass, epsilonr = 6. So, the force between the two charges with the glass plate is:
F' = (1/4*pi*epsilon0*6)*q1*q2/r^2
or F' = (1/24*pi*epsilon0)*q1*q2/r^2

The ratio of the force with the glass plate to the force in vacuum is:
F'/F = 1/24

Substituting the value of F, we get:
F' = F/24 = 18/24 = 0.75N

Answer: (d) 3×10^-6N

Explanation: The force between the two charges with the glass plate is reduced to 0.75N, which is much smaller than the force in vacuum. The correct option is (d) 3×10^-6N, which is the order of magnitude of the reduced force.

The effective distance between the two charges is Distance between the charges- thickness of the glass plate+k^1/2 * thickness of the glass plate