Question

Suppose you wanted to increase the force between two point charges by a factor of 5. By what factor must you change the distance between them

Answer 1

Answer:

the distance between the two point charges will be reduced by a factor of  $\frac{1}{\sqrt{5} } = 0.447$

Explanation:

The force between two point charges is given by Coulomb's law;

$F = \frac{kQ_1Q_2}{r^2}$

where;

k is Coulomb's constant

Q₁ and Q₂ are two point charges

r is the distance between the two charges

From the equation above, the following relationship between force and distance can be deduced;

F₁r₁² = F₂r₂²

F₁r₁² = (5F₁)r₂²

r₁² = 5r₂²

$r_2^2 = \frac{r_1^2}{5} \\\\r_2 = \sqrt{\frac{r_1^2}{5}} \\\\r_2 = \frac{r_1}{\sqrt{5} } \\\\r_2 = \frac{1}{\sqrt{5} } \ r_1\\\\r_2 = 0.447( r_1)$

Thus, the distance between the two point charges will be reduced by a factor of  $\frac{1}{\sqrt{5} } = 0.447$

Answer 2

To increase the force between two point charges by a factor of 5, you have to change the distance between them by a factor of 0.447.

Coulomb law states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the distance between them. It is given by:

F = kq₁q₂/r²

q₁, q₂ are charges, k is a constant and r is the distance between them.

Let us assume that:

F = kq₁q₂/r²

$r=\sqrt{\frac{kq_1q_2}{F} }$

To increase the force by a factor of 5:

$r=\sqrt{\frac{kq_1q_2}{5F} }\\\\r=0.447 \sqrt{\frac{kq_1q_2}{F} }$

Hence to increase the force between two point charges by a factor of 5, you have to change the distance between them by a factor of 0.447.

Find out more at: https://brainly.com/question/12819163