Two charges q1 and q2 are placed in vacuum at a distance d and force between themis F. If a medium of relative permittivity 4 is introduced between them then newforce will be:

Two charges q1 and q2 are placed in vacuum at a distance d and the force acting between them is F. If a medium of dielectric constant 4 is introduced around them, the force now will be4F2FF/2F/4

(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.

two-point charges placed at a distance 'r' in air exert a force 'F'. The distance at which they exert same force when placed in a certain medium (dielectric constant K) is :-

Coulomb's law for the magnitude of the force F𝐹 between two particles with charges  Q𝑄 and  Q′𝑄′ separated by a distance d𝑑 is|F|=K|QQ′|d2|𝐹|=𝐾|𝑄𝑄′|𝑑2,where K=14πϵ0𝐾=14𝜋𝜖0, and ϵ0=8.854×10−12C2/(N⋅m2)𝜖0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.Consider two point charges located on the x axis: one charge, q1𝑞1 = -19.5 nCnC , is located at x1𝑥1 = -1.690 mm ; the second charge, q2𝑞2 = 31.0 nCnC , is at the origin (x𝑥 = 0).Part AWhat is (Fnet3)x(𝐹net3)𝑥 , the x-component of the net force exerted by these two charges on a third charge q3𝑞3 = 48.5 nCnC placed between  q1𝑞1 and  q2𝑞2 at x3𝑥3 = -1.185 mm ?Your answer may be positive or negative, depending on the direction of the force.Express your answer numerically in newtons to three significant figures.View Available Hint(s)

Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomesF