(1) Two charges 10nC & 20nC are placed at 5m apart. Calculate (a) force between the charges, (b) force between the charges if a medium of relative permittivity 5 is introduced, (c) change in force between the charges and (d) force between the charges when they are brought in contact and then replaced in their initial positions.

Sure, let's solve this step by step:

(a) Force between the charges:

We can use Coulomb's law to calculate the force between two charges. The formula is:

F = k * |q1*q2| / r^2

where: F is the force between the charges, k is Coulomb's constant (9 * 10^9 N m^2/C^2), q1 and q2 are the charges (in this case, 10nC and 20nC, but we need to convert them to Coulombs, so it's 10 * 10^-9 C and 20 * 10^-9 C), r is the distance between the charges (5m).

So, F = 9 * 10^9 * (10 * 10^-9 * 20 * 10^-9) / (5^2) F = 7.2 N

(b) Force between the charges if a medium of relative permittivity 5 is introduced:

The force in a medium with relative permittivity (also known as dielectric constant) εr is given by:

F' = F / εr

So, F' = 7.2 / 5 = 1.44 N

(c) Change in force between the charges:

The change in force is simply the difference between the initial force and the force in the medium:

ΔF = F - F' = 7.2 - 1.44 = 5.76 N

(d) Force between the charges when they are brought in contact and then replaced in their initial positions:

When the charges are brought into contact, they will redistribute and the total charge will be the sum of the two charges. When they are replaced in their initial positions, the force between them will be:

F'' = k * (q1+q2)^2 / r^2

So, F'' = 9 * 10^9 * ((10 * 10^-9 + 20 * 10^-9)^2) / (5^2) F'' = 21.6 N