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 $q_{\text{red}}$.

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 $F$.

Goal:

We need to find the charge $q_{\text{red}}$ on the red sphere, given that the yellow sphere has charge $2q$.

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 $F$.

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.

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: $F_{\text{yellow}, y} = 2 k_e \frac{q^2}{d_2^2} \sin \theta$

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 $q_{\text{red}}$: $q_{\text{red}} = 2q \cdot \frac{d_1^2}{d_2^2 \cos \theta}$

Step 4: Final Expression for $q_{\text{red}}$

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

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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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?

Tip:

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