Question

the force between two points charges in air is 100N if the distance between them is incresed by 50% then the force between two charges will ne nearly equal to​

Answer 1

when then distance increased by 50%

then new distance will be 3/2r

as we know that Fα1/r²

then new force F'(we assume)

thenF/F'=(1/r²)÷[1/(3/2r)²]

100/F'=9/4

Then 9F'=400

then F'=44.44N