1) The electric force changes by a factor of 25
2) The electric force changes by a factor of 16/9
Explanation:
1)
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, let's call F the initial force between the two charges when they are at a distance of r.
Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is
r' = 5r
Therefore, the new force between the charges is:

So, the force has changed by a factor of 25.
2)
The original force between the two charges is
In this problem, we have:
- The distance between the charges is changed by a factor of 6:
r' = 6r
- The charges are both changed by a factor of 8:


Substituting into the equation, we find the new force:

So, the force has changed by a factor of 16/9.
Learn more about electric force:
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