Let's address the problems step-by-step. Specifically, we'll focus on question 5 which asks for the distance between two charges given their magnitudes and the force acting between them.
Firstly, the problem provides the following data:
1. Charge 1: [tex]\( Q_1 = 2 \times 10^{-7} \, \text{C} \)[/tex]
2. Charge 2: [tex]\( Q_2 = 4.5 \times 10^{-7} \, \text{C} \)[/tex]
3. Force between the charges: [tex]\( F = 0.1 \, \text{N} \)[/tex]
4. Coulomb’s constant: [tex]\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)[/tex]
The formula for the electrostatic force between two point charges is given by Coulomb's Law:
[tex]\[ F = k \frac{Q_1 Q_2}{r^2} \][/tex]
Here, we are tasked with finding the distance [tex]\( r \)[/tex] between the charges. Let's isolate [tex]\( r \)[/tex] in the equation.
1. Rearrange Coulomb's Law to solve for [tex]\( r^2 \)[/tex]:
[tex]\[
r^2 = k \frac{Q_1 Q_2}{F}
\][/tex]
2. Substitute the values given into the equation:
[tex]\[
r^2 = \left(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\right) \times \left(\frac{2 \times 10^{-7} \, \text{C} \times 4.5 \times 10^{-7} \, \text{C}}{0.1 \, \text{N}}\right)
\][/tex]
3. This evaluates to:
[tex]\[
r^2 = \left(8.99 \times 10^9 \right) \times \left(\frac{9 \times 10^{-14}}{0.1}\right)
\][/tex]
[tex]\[
r^2 = \left(8.99 \times 10^9 \right) \times 9 \times 10^{-13}
\][/tex]
4. Simplifying further, we get:
[tex]\[
r^2 = 8.091 \times 10^{-3} \, \text{m}^2
\][/tex]
5. Now, take the square root of both sides to find [tex]\( r \)[/tex]:
[tex]\[
r = \sqrt{8.091 \times 10^{-3}}
\][/tex]
6. Therefore:
[tex]\[
r \approx 0.08995 \, \text{m}
\][/tex]
Thus, the distance [tex]\( r \)[/tex] between the two charges is approximately [tex]\( 0.08995 \, \text{m} \)[/tex].
In summary, given the magnitudes of the charges and the force between them, the distance between the charges is approximately [tex]\( 0.08995 \, \text{m} \)[/tex], or about [tex]\( 0.09 \, \text{m} \)[/tex].
