Question

What is the force of repulsion between two charges of 1 coulomb is kept 1 metre apart in vacuum?

Answer 1

Let the 2 charges be q1 q2

Fi be Force in initial conditions

Ff be force in final conditions

r be final distance

Assuming you know K=1/(4πε0) and K′=1/(4πε0∗εr)

Here K is for vacuum with absolute permittivity ε0 and K’ is for other medium with relative permittivity εr (in our case oil with value 4)

Applying coulomb’s law on both conditions we get,

Fi=Kq1∗q2/12

Ff=K′q1∗q2/r2

Ff=(K/4)∗q1∗q2/r2

Given

Fi=Ff

Fi=Fi/4∗r2

Solving this we get,

4∗r2=1

Hence

r=1/2m=50cm.

Answer 2

answer is already given bro