Electrostatic force between two identical charges placed in vacuum at distance of r is F . A slab of width r 5 and dielectric constant 9 is inserted between these two charges, then the force between the charges is

Step-by-step solution

Force between the two charges when it is placed in vacuum and distance between them is r is F = 1 4 π ε 0 q 1 q 2 r 2 . Now, the force when a dielectric k is inserted between these two charges is F = 1 4 π ε 0 q 1 q 2 k r ' 2 . On comparing the dielectric force with the vacuum force we will get, r 2 = k r ' 2 Or r ' = r k Hence, the effective thickness of slab is r 5 k . Thus, the force between the charges is F ' = 1 4 π ε 0 q 1 q 2 r 5 + r 5 9 2 = q 1 q 2 4 π ε 0