Question

Can someone solve this problem and explain to me how you got it​

Answer 1

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:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ 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:

$F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}$

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

$F=k\frac{q_1 q_2}{r^2}$

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:

$q_1' = 8q_1$

$q_2' = 8q_2$

Substituting into the equation, we find the new force:

$F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F$

So, the force has changed by a factor of 16/9.

Learn more about electric force:

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