Question

Force between two identical charges placed at a distance r in vacuum is f now a slab of dielectric

Answer 1

Hey Dear,

◆ Complete question is -

Q. Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric constant K=4 is inserted between these two charges. The thickness of the slab is r/2. The force between the charges will now become ___.

● Answer -

F' = F

◆ Explanation -

Electrostatic force between two charges in vacuum is given by -

F = (1/4πε0) × q1.q2 / r²

F = (1/4πε0) × q² / r²

Electrostatic force between two charges in medium is given by -

F' = (1/4πε0K) × q² / (r/2)²

F' = [(1/4πε0)q²/r²] × 4/K

F' = F × 4 / 4

F' = F

Therefore, the force between two charges will remain same after putting slab.

Hope this helps you. Thanks for asking.