Question

consider the following configuration of fixed, uniformly charged spheres in (figure 1).

• a blue sphere fixed at the origin with positive charge...

figure (shows blue sphere at origin, red sphere at (d₁,0) on x - axis, yellow sphere at distance d₂ from origin with angle θ between blue - yellow line and x - axis)
part c
suppose that the magnitude of the charge on the yellow sphere is determined to be 2q. calculate the charge ( q_{\text{red}} ) on the red sphere.
express your answer in terms of ( q ), ( d_1 ), ( d_2 ), and ( \theta ).
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( q_{\text{red}} = )
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Explanation:

d_2 = 0\), which is not the case. Therefore, my mistake must be in the charge signs. Maybe the blue sphere is positive, yellow is negative, and red is negative? Wait, let's start over.

Let’s denote:

• Blue sphere: charge \(q\) (positive), at origin.

• Red sphere: charge \(q_{red}\), at \((d_1, 0)\).

• Yellow sphere: charge \(2q\) (let's assume it's negative, so force on red from yellow is attraction, and force from blue is repulsion or attraction depending on \(q_{red}\)).

The force on red from blue: \( \vec{F_{br}} = k\frac{q q_{red}}{d_1^2} \hat{i} \) (if \(q_{red}\) is positive, repulsion, +x; if negative, attraction, -x).

The force on red from yellow: \( \vec{F_{yr}} = k\frac{(2q) q_{red}}{r^2} \vec{r_{yellow - red}} / r \), where \( \vec{r_{yellow - red}} = (d_1 - d_2\cos\theta, d_2\sin\theta) \), so \( \vec{F_{yr}} = k\frac{2q q_{red}}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) \).

For equilibrium, \( \vec{F_{br}} + \vec{F_{yr}} = 0 \), so:

\( k\frac{q q_{red}}{d_1^2} + k\frac{2q q_{red}}{r^3}(d_1 - d_2\cos\theta) = 0 \) (x - component)

\( k\frac{2q q_{red}}{r^3}(d_2\sin\theta) = 0 \) (y - component)

From the y - component: \( d_2\sin\theta = 0 \), which is impossible unless \(\theta = 0\) or \(d_2 = 0\). Therefore, the only way this works is if the yellow sphere is arranged such that the y - component of the force on red is zero, which means that the yellow sphere is on the x - axis, so \(\theta = 0\) or \(\theta = 180^\circ\), but the figure shows \(\theta\) is not zero. Wait, maybe the red sphere is in equilibrium under the forces from blue and yellow, and the yellow sphere's force on red has no y - component, which would mean that the yellow sphere is on the x - axis, so \(\theta = 0\) (yellow to the right of blue) or \(\theta = 180^\circ\) (yellow to the left). But the figure shows yellow at an angle \(\theta\) below the x - axis. This suggests that maybe the blue sphere, red sphere, and yellow sphere form a triangle where the force on red from blue and yellow are balanced, and the y - component of the yellow force is balanced by... but all spheres are fixed, so the red sphere is in equilibrium, so the net force is zero. Therefore, the y - component must be zero, so \(d_2\sin\theta = 0\), which is a contradiction. Therefore, my initial assumption about the charge signs is wrong. Maybe the blue sphere is positive, yellow is positive, and red is negative. Then:

\( \vec{F_{br}} = k\frac{q (-q_{red})}{d_1^2} \hat{i} = -k\frac{q q_{red}}{d_1^2} \hat{i} \) (attraction, -x direction)

\( \vec{F_{yr}} = k\frac{2q (-q_{red})}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) = -k\frac{2q q_{red}}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) \) (attraction, towards yellow, so the vector is from red to yellow, which is \((d_2\cos\theta - d_1, -d_2\sin\theta)\), so I had the direction wrong earlier. The force on red from yellow is towards yellow, so \( \vec{F_{yr}} = k\frac{2q q_{red}}{r^2} \frac{\vec{r_{blue - yellow}} - \vec{r_{blue - red}}}{r} = k\frac{2q q_{red}}{r^3} (d_2\cos\theta - d_1, -d_2\sin\theta) \)

Now, the x - component of \(F_{yr}\) is \( k\frac{2q q_{red}}{r^3} (d_2\cos\theta - d_1) \)

The y - component is \( k\frac{2q q_{red}}{r^3} (-d_2\sin\theta) \)

For equilibrium, net force is zero:

\( -k\frac{q q_{red}}{d

Answer:

d_2 = 0\), which is not the case. Therefore, my mistake must be in the charge signs. Maybe the blue sphere is positive, yellow is negative, and red is negative? Wait, let's start over.

Let’s denote:

• Blue sphere: charge \(q\) (positive), at origin.

• Red sphere: charge \(q_{red}\), at \((d_1, 0)\).

• Yellow sphere: charge \(2q\) (let's assume it's negative, so force on red from yellow is attraction, and force from blue is repulsion or attraction depending on \(q_{red}\)).

The force on red from blue: \( \vec{F_{br}} = k\frac{q q_{red}}{d_1^2} \hat{i} \) (if \(q_{red}\) is positive, repulsion, +x; if negative, attraction, -x).

The force on red from yellow: \( \vec{F_{yr}} = k\frac{(2q) q_{red}}{r^2} \vec{r_{yellow - red}} / r \), where \( \vec{r_{yellow - red}} = (d_1 - d_2\cos\theta, d_2\sin\theta) \), so \( \vec{F_{yr}} = k\frac{2q q_{red}}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) \).

For equilibrium, \( \vec{F_{br}} + \vec{F_{yr}} = 0 \), so:

\( k\frac{q q_{red}}{d_1^2} + k\frac{2q q_{red}}{r^3}(d_1 - d_2\cos\theta) = 0 \) (x - component)

\( k\frac{2q q_{red}}{r^3}(d_2\sin\theta) = 0 \) (y - component)

From the y - component: \( d_2\sin\theta = 0 \), which is impossible unless \(\theta = 0\) or \(d_2 = 0\). Therefore, the only way this works is if the yellow sphere is arranged such that the y - component of the force on red is zero, which means that the yellow sphere is on the x - axis, so \(\theta = 0\) or \(\theta = 180^\circ\), but the figure shows \(\theta\) is not zero. Wait, maybe the red sphere is in equilibrium under the forces from blue and yellow, and the yellow sphere's force on red has no y - component, which would mean that the yellow sphere is on the x - axis, so \(\theta = 0\) (yellow to the right of blue) or \(\theta = 180^\circ\) (yellow to the left). But the figure shows yellow at an angle \(\theta\) below the x - axis. This suggests that maybe the blue sphere, red sphere, and yellow sphere form a triangle where the force on red from blue and yellow are balanced, and the y - component of the yellow force is balanced by... but all spheres are fixed, so the red sphere is in equilibrium, so the net force is zero. Therefore, the y - component must be zero, so \(d_2\sin\theta = 0\), which is a contradiction. Therefore, my initial assumption about the charge signs is wrong. Maybe the blue sphere is positive, yellow is positive, and red is negative. Then:

\( \vec{F_{br}} = k\frac{q (-q_{red})}{d_1^2} \hat{i} = -k\frac{q q_{red}}{d_1^2} \hat{i} \) (attraction, -x direction)

\( \vec{F_{yr}} = k\frac{2q (-q_{red})}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) = -k\frac{2q q_{red}}{r^3} (d_1 - d_2\cos\theta, d_2\sin\theta) \) (attraction, towards yellow, so the vector is from red to yellow, which is \((d_2\cos\theta - d_1, -d_2\sin\theta)\), so I had the direction wrong earlier. The force on red from yellow is towards yellow, so \( \vec{F_{yr}} = k\frac{2q q_{red}}{r^2} \frac{\vec{r_{blue - yellow}} - \vec{r_{blue - red}}}{r} = k\frac{2q q_{red}}{r^3} (d_2\cos\theta - d_1, -d_2\sin\theta) \)

Now, the x - component of \(F_{yr}\) is \( k\frac{2q q_{red}}{r^3} (d_2\cos\theta - d_1) \)

The y - component is \( k\frac{2q q_{red}}{r^3} (-d_2\sin\theta) \)

For equilibrium, net force is zero:

\( -k\frac{q q_{red}}{d