Two point charges are placed on the x-axis as follows:

Charge q_(1)=+4.00nC is located at x=0.200m, and charge q_(2)=+6.00nC is at x=-0.500m.

What are the magnitude and direction of the total force exerted by these two charges on a

negative point charge q_(3)=-6.00nC that is placed at the origin?

(note: 1nC=1 nanocoulomb =1\times 10^(-9)C )

k=8.99\times 10^(9)N(m^(2))/(C^(2))

To solve this problem, we'll calculate the forces exerted on charge \( q_3 \) by both \( q_1 \) and \( q_2 \) using Coulomb's law, and then find the vector sum of these forces.

### Step 1: Coulomb’s Law
Coulomb's law is given by:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where:
- \( F \) is the magnitude of the force between the charges,
- \( k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step 2: Calculate Forces Exerted by \( q_1 \) and \( q_2 \) on \( q_3 \)

#### Force exerted by \( q_1 \) on \( q_3 \):
- \( q_1 = +4.00 \, \text{nC} = 4.00 \times 10^{-9} \, \text{C} \),
- \( q_3 = -6.00 \, \text{nC} = -6.00 \times 10^{-9} \, \text{C} \),
- Distance between \( q_1 \) (at \( x = 0.200 \, \text{m} \)) and \( q_3 \) (at the origin) is \( r_1 = 0.200 \, \text{m} \).

Using Coulomb's law:
\[
F_{13} = k \frac{|q_1 q_3|}{r_1^2} = 8.99 \times 10^9 \frac{|(4.00 \times 10^{-9})(-6.00 \times 10^{-9})|}{(0.200)^2}
\]
\[
F_{13} = 8.99 \times 10^9 \times \frac{24.00 \times 10^{-18}}{0.0400} = 5.394 \times 10^{-7} \, \text{N}
\]
Since \( q_1 \) is positive and \( q_3 \) is negative, the force will be attractive, so \( q_3 \) will be pulled towards \( q_1 \). The force is directed to the right (positive \( x \)-axis).

#### Force exerted by \( q_2 \) on \( q_3 \):
- \( q_2 = +6.00 \, \text{nC} = 6.00 \times 10^{-9} \, \text{C} \),
- Distance between \( q_2 \) (at \( x = -0.500 \, \text{m} \)) and \( q_3 \) (at the origin) is \( r_2 = 0.500 \, \text{m} \).

Using Coulomb's law:
\[
F_{23} = k \frac{|q_2 q_3|}{r_2^2} = 8.99 \times 10^9 \frac{|(6.00 \times 10^{-9})(-6.00 \times 10^{-9})|}{(0.500)^2}
\]
\[
F_{23} = 8.99 \times 10^9 \times \frac{36.00 \times 10^{-18}}{0.2500} = 1.295 \times 10^{-7} \, \text{N}
\]
Again, since \( q_2 \) is positive and \( q_3 \) is negative, the force will be attractive, so \( q_3 \) will be pulled towards \( q_2 \). The force is directed to the left (negative \( x \)-axis).

### Step 3: Calculate the Net Force

Now, we calculate the net force by adding the forces as vectors. The force exerted by \( q_1 \) pulls to the right (positive \( x \)-axis) and the force exerted by \( q_2 \) pulls to the left (negative \( x \)-axis).

Let:
- \( F_{13} \) be the force due to \( q_1 \) acting to the right (positive),
- \( F_{23} \) be the force due to \( q_2 \) acting to the left (negative).

The net force is:
\[
F_{\text{net}} = F_{13} - F_{23} = (5.394 \times 10^{-7}) - (1.295 \times 10^{-7}) = 4.099 \times 10^{-7} \, \text{N}
\]
The positive result indicates that the net force is directed to the right (towards \( q_1 \)).

### Step 4: Final Answer
- **Magnitude of the net force**: \( 4.099 \times 10^{-7} \, \text{N} \),
- **Direction**: To the right (positive \( x \)-axis).

Would you like further details or have any questions?

---

Here are five related questions:
1. How does the force change if \( q_3 \) were positive instead of negative?
2. How does the distance between charges affect the magnitude of the force?
3. What happens if \( q_2 \) is moved closer to the origin?
4. Can you explain how the direction of force is determined in such problems?
5. How would the net force change if the charge \( q_1 \) were doubled?

**Tip**: Remember that Coulomb's law force decreases with the square of the distance between charges.

Two point charges are placed on the x-axis as follows:

Charge q1=+4.00nC is located at x=0.200m, and charge q2=+6.00nC is at x=-0.500m.

What are the magnitude and direction of the total force exerted by these two charges on a negative point charge q3=-6.00nC that is placed at the origin?

(note: 1nC=1 nanocoulomb =1×10^(-9)C)

k=8.99×10^9N(m^2)/(C^2)

Calculate the magnitude and direction of the net force exerted by two point charges on a negative charge placed at the origin using Coulomb’s law. Includes a detailed solution with vector analysis and force calculations, covering core electrostatic concepts. Ideal for students in Grades 11-12.

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