1) The electric force has changed by a factor of 40.
2) The electric force has changed by a factor of 25.
Explanation:
1)
The magnitude of the electric 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
Let's call F the initial electric force between the two charges in this problem. Later, the charge of one of the particles is changed by a factor of 40, therefore
[tex]q_1' = 40 q_1[/tex]
(we have assumed that the charge has been increased rather than decreased, since it is not specified)
Substituting into the equation, we can find the new electric force:
[tex]F'=k\frac{q_1' q_2}{r^2}=k\frac{(40q_1) q_2}{r^2}=40(k\frac{q_1 q_2}{r^2})=40F[/tex]
Therefore, the force has changed by a factor of 40.
2)
Let's now call again the initial electric force between the two charges F:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
In this second problem, the charge of both particles has been changed by a factor 5, therefore
[tex]q_1' = 5 q_1[/tex]
[tex]q_2' = 5q_2[/tex]
(again, we have assumed that the charge has been increased rather than decreased, since it is not specified)
Substituting into the equation for F, we can find the new electric force between the two particles:
[tex]F'=k\frac{q_1' q_2'}{r^2}=k\frac{(5q_1) (5q_2)}{r^2}=25(k\frac{q_1 q_2}{r^2})=25F[/tex]
Therefore, the force has changed by a factor of 25.
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