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Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. Calculate the charge qred on the red sphere based on this image.

Thank you for the image. Let's break down the information step by step and proceed with solving for the charge on the red sphere \( q_{\text{red}} \).

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### Problem Setup:
- **Blue sphere**: Located at the origin \((0,0)\) with charge \( q \).
- **Red sphere**: Located at \( (d_1, 0) \) with unknown charge \( q_{\text{red}} \).
- **Yellow sphere**: Located at \( (d_2 \cos \theta, -d_2 \sin \theta) \) with charge \( q_{\text{yellow}} = 2q \).
- **Net electric force** on the blue sphere is **in the \(-y\) direction** with magnitude \( F \).

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### Goal:
We need to find the charge \( q_{\text{red}} \) on the red sphere, given that the yellow sphere has charge \( 2q \).

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### Strategy:

Since the **net force on the blue sphere is along the \(-y\)-axis**, the forces from the red and yellow spheres must balance in the **horizontal \(x\)-direction**, while the net force in the \(y\)-direction comes only from the yellow sphere. This balance gives us two key conditions:
1. **Horizontal force** (from the red and yellow charges) must cancel.
2. **Vertical force** comes from the yellow charge alone, which must equal the given magnitude \( F \).

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### Step 1: Coulomb's Law

The magnitude of the force between two point charges is given by:
\[
F = k_e \frac{|q_1 q_2|}{r^2}
\]
where \( k_e \) is Coulomb's constant.

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### Step 2: Forces on the Blue Sphere

1. **Force from the Red Sphere** (along \( +x \)-direction):
   - Distance: \( d_1 \).
   - Force magnitude:
     \[
     F_{\text{red}} = k_e \frac{q \cdot q_{\text{red}}}{d_1^2}
     \]
   - Direction: Positive \(x\)-axis.

2. **Force from the Yellow Sphere**:
   - Distance: \( d_2 \).
   - Force magnitude:
     \[
     F_{\text{yellow}} = k_e \frac{q \cdot (2q)}{d_2^2} = 2 k_e \frac{q^2}{d_2^2}
     \]
   - Components:
     - Horizontal component: \( F_{\text{yellow}, x} = 2 k_e \frac{q^2}{d_2^2} \cos \theta \)
     - Vertical component: \( F_{\text{yellow}, y} = 2 k_e \frac{q^2}{d_2^2} \sin \theta \)

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### Step 3: Force Balance in the Horizontal Direction

Since the net force on the blue sphere is only in the \(-y\)-direction, the horizontal forces must cancel:
\[
F_{\text{red}} = F_{\text{yellow}, x}
\]
Substitute the expressions:
\[
k_e \frac{q \cdot q_{\text{red}}}{d_1^2} = 2 k_e \frac{q^2}{d_2^2} \cos \theta
\]
Simplify:
\[
\frac{q \cdot q_{\text{red}}}{d_1^2} = 2 \frac{q^2}{d_2^2} \cos \theta
\]
Solve for \( q_{\text{red}} \):
\[
q_{\text{red}} = 2q \cdot \frac{d_1^2}{d_2^2 \cos \theta}
\]

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### Step 4: Final Expression for \( q_{\text{red}} \)

The charge on the red sphere is:
\[
q_{\text{red}} = 2q \cdot \frac{d_1^2}{d_2^2 \cos \theta}
\]

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### Conclusion

The charge on the red sphere, \( q_{\text{red}} \), is proportional to \( 2q \) and depends on the distances \( d_1 \) and \( d_2 \), as well as the angle \( \theta \). 

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Would you like to see further details or discuss a specific part of this solution?

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### Related Questions:
1. How does the angle \( \theta \) affect the force balance?
2. What would happen if the charges on the red and yellow spheres were reversed?
3. How does the net force in the \(y\)-direction change if \( \theta = 0^\circ \)?
4. What is the significance of Coulomb’s constant \( k_e \) in the calculations?
5. Can we determine the sign of \( q_{\text{red}} \) from the direction of forces?

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### Tip:
Always double-check if the directions of forces align with your assumptions when working with vector quantities in electric force problems!

Learn how to solve for the charge on the red sphere (qred) in an electrostatics problem involving three charged spheres. The problem uses Coulomb's Law, vector components, and the principle of superposition. Suitable for undergraduate physics students.

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