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

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"  

 

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$  (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:

$F_{E}= K\frac{6 q_{1}. 6 q_{2}}{(6d)^{2}}$  (2)

$F_{E}= 36 K\frac{q_{1}. q_{2}}{36d^{2}}$  (3)

Simplifying:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$  (4)

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