To solve for the initial speed of the electron, we will use the concept of energy conservation. Specifically, the total mechanical energy of the electron is conserved as it moves under the influence of the electric field created by the charged sphere.

Steps:

1. Electric potential energy: The electron starts far away, where the electric potential energy is nearly zero. As it moves closer to the charged sphere, its electric potential energy increases due to the charge on the sphere.

2. Kinetic energy: The electron initially has only kinetic energy, and as it moves closer to the sphere, some of this energy is converted into electric potential energy until it stops and turns around.

At the point closest to the sphere (0.31 mm away from the surface), the electron's kinetic energy becomes zero, and all its energy is in the form of electric potential energy.

Key formulas:

• The initial kinetic energy of the electron is given by: $KE_{\text{initial}} = \frac{1}{2} m_e v_0^2$ where $m_e = 9.11 \times 10^{-31} \, \text{kg}$ is the mass of the electron and $v_0$ is the initial speed we are solving for.

• The electric potential energy at distance $r$ from a charged sphere is: $U = \frac{k_e Q q}{r}$ where:

  • $k_e = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$ (Coulomb's constant),
  • $Q = -4.4 \times 10^{-9} \, \text{C}$ (charge on the sphere),
  • $q = -1.6 \times 10^{-19} \, \text{C}$ (charge of the electron),
  • $r$ is the distance between the electron and the center of the sphere.

Distance from the center of the sphere:

The total distance $r$ from the center of the sphere to the electron at the closest point is: $r = R_{\text{sphere}} + 0.31 \, \text{mm}$ where $R_{\text{sphere}}$ is the radius of the sphere: $R_{\text{sphere}} = \frac{2.3 \, \text{mm}}{2} = 1.15 \, \text{mm} = 1.15 \times 10^{-3} \, \text{m}$ So the total distance $r$ is: $r = 1.15 \times 10^{-3} \, \text{m} + 0.31 \times 10^{-3} \, \text{m} = 1.46 \times 10^{-3} \, \text{m}$

Energy conservation equation:

At the closest point, the electron’s kinetic energy is converted into electric potential energy: $\frac{1}{2} m_e v_0^2 = \frac{k_e Q q}{r}$

Now let's plug in the values and solve for the initial speed $v_0$.

$v_0 = \sqrt{\frac{2 k_e Q q}{m_e r}}$

Substitute the known values:

• $k_e = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$,
• $Q = -4.4 \times 10^{-9} \, \text{C}$,
• $q = -1.6 \times 10^{-19} \, \text{C}$,
• $m_e = 9.11 \times 10^{-31} \, \text{kg}$,
• $r = 1.46 \times 10^{-3} \, \text{m}$.

I'll calculate the initial speed now.The initial speed of the electron is approximately $9.76 \times 10^7 \, \text{m/s}$.

Would you like more details on any part of this process, or have any questions?

Here are 5 related questions for further exploration:

1. How does the charge of the sphere affect the initial speed of the electron?
2. What happens to the electron's speed if the distance to the sphere is increased?
3. How would the result change if the sphere had a positive charge instead of a negative charge?
4. What is the relationship between the initial kinetic energy and the potential energy at the closest point?
5. How can we compute the time taken by the electron to reach the closest point?

Tip: The principle of energy conservation is powerful in physics, especially when dealing with fields and forces like gravity or electromagnetism.