Question

According to coulombs law, what will happen to the force between two charged particles if the magnitude are increased by 6 times and the distance between them is increased by 6 times

Answer 1

Answer: The electrostatic force will be the same

Step-by-step explanation:

According to Coulomb's Law, when two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies or particles.

In this sense, this law states the following:

"The electrostatic force
`F_(E)` between two point charges
`q_(1)` and
`q_(2)` is proportional to the product of the charges and inversely proportional to the square of the distance
`d` that separates them, and has the direction of the line that joins them"

`F_(E)= K(q_(1).q_(2))/(d^(2))` (1)

Being
`K` is a proportionality constant.

Now, if each
`q_(1)` and
`q_(2)` are increased by 6, and the distance between them as well, we will have the following:

`F_(E)= K(6 q_(1). 6 q_(2))/((6d)^(2))` (2)

`F_(E)= 36 K(q_(1). q_(2))/(36d^(2))` (3)

Simplifying:

`F_(E)= K(q_(1).q_(2))/(d^(2))` (4)

Comparing (1) with (4) we can see the electrostatic force is the same.