We can solve this problem using Coulomb's law, which gives the electrostatic force between two point charges. The formula is:

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

Where:

• $F$ is the electrostatic force between the charges (7.2 N),
• $k$ is Coulomb's constant, $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$,
• $q_1$ and $q_2$ are the charges on the spheres (which are equal, so we can just call them $q$),
• $r$ is the distance between the charges (0.20 meters).

Since $q_1 = q_2 = q$, we can simplify the formula as:

$F = k \frac{q^2}{r^2}$

Now, we can solve for $q$:

$q^2 = \frac{F r^2}{k}$

Substitute the known values:

$q^2 = \frac{(7.2 \, \text{N})(0.20^2 \, \text{m}^2)}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2}$

$q^2 = \frac{7.2 \times 0.04}{8.99 \times 10^9}$

$q^2 = \frac{0.288}{8.99 \times 10^9}$

$q^2 = 3.21 \times 10^{-11} \, \text{C}^2$

$q = \sqrt{3.21 \times 10^{-11}} \, \text{C}$

$q \approx 5.66 \times 10^{-6} \, \text{C}$

So, the magnitude of each charge is approximately 5.66 µC (microcoulombs).

Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions:

1. How does the electrostatic force change if the distance between the charges is doubled?
2. What would the force be if the charges were of opposite signs instead of the same sign?
3. How does the magnitude of Coulomb's constant $k$ affect the electrostatic force?
4. If the charges were 0.5 meters apart, how would you adjust the calculation for $q$?
5. How can Coulomb's law be applied to real-world objects, such as charged balloons?

Tip: Remember that Coulomb’s law is only accurate for point charges or very small spherical objects where the charge is evenly distributed!