To determine the direction of the net electric force on each sphere due to the other two, we can break down the problem step by step.

1. Calculate the Force Between Each Pair of Spheres:

Using Coulomb's law, the magnitude of the electric force between two point charges is given by:

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

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 (11 cm = 0.11 m).

Forces between Spheres 1 and 2:

Charges: $q_1 = 31 \, \mu\text{C} = 31 \times 10^{-6} \, \text{C}$ and $q_2 = -50 \, \mu\text{C} = -50 \times 10^{-6} \, \text{C}$

$F_{12} = \frac{8.99 \times 10^9 \times |31 \times 10^{-6} \times -50 \times 10^{-6}|}{(0.11)^2}$

Forces between Spheres 1 and 3:

Charges: $q_1 = 31 \, \mu\text{C} = 31 \times 10^{-6} \, \text{C}$ and $q_3 = 46 \, \mu\text{C} = 46 \times 10^{-6} \, \text{C}$

$F_{13} = \frac{8.99 \times 10^9 \times |31 \times 10^{-6} \times 46 \times 10^{-6}|}{(0.22)^2}$

Here $r = 22$ cm or 0.22 m because sphere 1 is at one end and sphere 3 at the other.

Forces between Spheres 2 and 3:

Charges: $q_2 = -50 \, \mu\text{C} = -50 \times 10^{-6} \, \text{C}$ and $q_3 = 46 \, \mu\text{C} = 46 \times 10^{-6} \, \text{C}$

$F_{23} = \frac{8.99 \times 10^9 \times |50 \times 10^{-6} \times 46 \times 10^{-6}|}{(0.11)^2}$

2. Determine the Direction of Each Force:

• The force between sphere 1 and sphere 2 ($F_{12}$) is attractive because they have opposite charges. Sphere 1 is pulled toward sphere 2.
• The force between sphere 1 and sphere 3 ($F_{13}$) is repulsive because they have like charges. Sphere 1 is pushed away from sphere 3.
• The force between sphere 2 and sphere 3 ($F_{23}$) is attractive because they have opposite charges. Sphere 2 is pulled toward sphere 3.

3. Determine the Net Force on Each Sphere:

To find the net force on each sphere, sum the forces acting on it, taking into account the directions (left/right):

• Sphere 1:

  • Attracted to sphere 2 (left).
  • Repelled by sphere 3 (right).

  The net force on sphere 1 will be the vector sum of these two forces.

• Sphere 2:

  • Attracted to sphere 1 (right).
  • Attracted to sphere 3 (right).

  The net force on sphere 2 will be the sum of these two attractive forces, both pulling to the right.

• Sphere 3:

  • Repelled by sphere 1 (left).
  • Attracted to sphere 2 (left).

  The net force on sphere 3 will be the vector sum of these two forces, which will both be directed to the left.

4. Conclusion:

• Sphere 1 experiences a net force to the left.
• Sphere 2 experiences a net force to the right.
• Sphere 3 experiences a net force to the left.

The direction of the net electric force on each sphere depends on the combination of attractive and repulsive forces exerted by the other two spheres.

Do you want details on any of these steps, or do you have any further questions?

Related Questions:

1. How would the net forces change if the distances between the spheres were doubled?
2. What happens to the net force on each sphere if the charge on one of the spheres is doubled?
3. How does the direction of force change if all spheres have the same charge sign?
4. How would introducing a fourth charged sphere affect the forces on these three spheres?
5. What is the potential energy of this system of three charges?

Tip: Always consider the vector nature of forces when analyzing electric forces in multiple charge systems.