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:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] 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:
[tex]F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}[/tex]
So, the force has changed by a factor of 25.
2)
The original force between the two charges is
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
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:
[tex]q_1' = 8q_1[/tex]
[tex]q_2' = 8q_2[/tex]
Substituting into the equation, we find the new force:
[tex]F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F[/tex]
So, the force has changed by a factor of 16/9.
Learn more about electric force:
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