To solve this problem, we need to use Coulomb's Law, which gives the magnitude of the electrostatic force between two charges:

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

where:

• $F$ = force between the charges (in Newtons, N)
• $q_1$ and $q_2$ = magnitudes of the charges (in Coulombs, C)
• $r$ = distance between the charges (in meters, m)
• $k$ = Coulomb's constant, which is $\frac{1}{4 \pi \epsilon_0}$, where $\epsilon_0$ is the permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$.

Given values:

• $q_1 = 95 \, \mu\text{C} = 95 \times 10^{-6} \, \text{C}$
• $q_2 = 135 \, \mu\text{C} = 135 \times 10^{-6} \, \text{C}$
• $r = 2.0 \, \text{m}$
• $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$

Now, calculate $k$: $k = \frac{1}{4 \pi \epsilon_0} = \frac{1}{4 \pi \times 8.85 \times 10^{-12}} \approx 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$

Substitute into Coulomb's Law: $F = \frac{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \cdot |95 \times 10^{-6} \, \text{C} \cdot 135 \times 10^{-6} \, \text{C}|}{(2.0 \, \text{m})^2}$

Let's calculate this value.The magnitude of the force exerted by the charge $q_1$ on $q_2$ is approximately $28.82 \, \text{N}$.

Would you like further details or have any questions?

Here are some related questions to explore further:

1. How does the force change if the distance between the charges is halved?
2. What would happen to the force if one of the charges was negative?
3. How is Coulomb's Law similar to Newton's law of universal gravitation?
4. What is the role of the permittivity of free space $\epsilon_0$ in this calculation?
5. How does the force change if the charges are placed in a medium other than a vacuum?

Tip: Remember that the direction of the force between two charges depends on whether the charges are like or unlike. Like charges repel each other, while unlike charges attract each other.