We are asked to find the force on the charge located at 8 cm (with charge $-2q$) due to the other two charges at 3 cm (+q) and 11 cm (+q).

Step 1: Identify the charges and positions

• Charge 1 ($q_1 = +q$) is at 3 cm.
• Charge 2 ($q_2 = -2q$) is at 8 cm (this is the charge on which we will calculate the force).
• Charge 3 ($q_3 = +q$) is at 11 cm.

Given that $q = 1.250 \, \mu C = 1.250 \times 10^{-6} \, C$.

Step 2: Use Coulomb's Law

Coulomb's law for the force between two point charges is given by:

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

where:

• $F$ is the magnitude of the force,
• $k_e = 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$ is Coulomb's constant,
• $q_1$ and $q_2$ are the charges,
• $r$ is the distance between the charges.

We need to calculate the force on $q_2 = -2q$ due to the other two charges $q_1$ and $q_3$.

Step 3: Forces between the charges

Force between $q_1$ and $q_2$:

The distance between $q_1$ and $q_2$ is $8 \, \text{cm} - 3 \, \text{cm} = 5 \, \text{cm} = 0.05 \, \text{m}$.

$F_{12} = k_e \frac{|q_1 q_2|}{r_{12}^2} = (8.988 \times 10^9) \frac{|(1.250 \times 10^{-6}) \times (-2 \times 1.250 \times 10^{-6})|}{(0.05)^2}$

$F_{12} = (8.988 \times 10^9) \frac{|(-2 \times 1.250^2 \times 10^{-12})|}{0.0025}$

$F_{12} = (8.988 \times 10^9) \frac{3.125 \times 10^{-12}}{0.0025}$

$F_{12} = 11.18 \, \text{N}$

Since the charges are opposite in sign, the force between them is attractive. This means $F_{12}$ will be directed to the left (toward $q_1$).

Force between $q_2$ and $q_3$:

The distance between $q_2$ and $q_3$ is $11 \, \text{cm} - 8 \, \text{cm} = 3 \, \text{cm} = 0.03 \, \text{m}$.

$F_{23} = k_e \frac{|q_2 q_3|}{r_{23}^2} = (8.988 \times 10^9) \frac{|(-2 \times 1.250 \times 10^{-6}) \times (1.250 \times 10^{-6})|}{(0.03)^2}$

$F_{23} = (8.988 \times 10^9) \frac{3.125 \times 10^{-12}}{0.0009}$

$F_{23} = 31.25 \, \text{N}$

Since both charges are of the same sign ($+q$ and $-2q$), the force between them is repulsive. This means $F_{23}$ will be directed to the right (away from $q_3$).

Step 4: Net Force on $q_2$

Now, we calculate the net force on $q_2$, which is the sum of the two forces $F_{12}$ and $F_{23}$.

• The force $F_{12}$ is to the left (negative direction).
• The force $F_{23}$ is to the right (positive direction).

Thus, the net force on $q_2$ is:

$F_{\text{net}} = F_{23} - F_{12} = 31.25 \, \text{N} - 11.18 \, \text{N} = 20.07 \, \text{N}$

The net force is directed to the right (positive direction).

Final Answer:

The net force on the charge located at 8 cm is $\mathbf{20.07 \, N}$ to the right.

Would you like further clarification on any step or have other questions?

Here are some related questions you might find interesting:

1. What would the net force be if all the charges had the same sign?
2. How would the force change if the charges were doubled in magnitude?
3. What happens to the force if the distance between the charges is halved?
4. How does the sign of a charge affect the direction of the force?
5. What is the significance of Coulomb's constant in this equation?

Tip: The force between charges decreases rapidly with increasing distance (since it is inversely proportional to the square of the distance), so even small changes in distance can have a big effect on the force.