Let's solve the problem step by step.

Given:

• Uniform electric field: $\vec{E} = -200 \, \text{N/C} \, \hat{j}$ (negative indicates downward direction).
• Initial velocity of the electron: $\vec{v}_0 = 3 \times 10^6 \, \text{m/s} \, \hat{i}$ (along $x$-direction).
• Length of the field region: $L = 0.1 \, \text{m}$.
• Charge of electron: $q = -1.6 \times 10^{-19} \, \text{C}$.
• Mass of electron: $m = 9.1 \times 10^{-31} \, \text{kg}$.

Part (a) Acceleration of the electron

The force on the electron due to the electric field is given by: $\vec{F} = q \vec{E}$ Substitute the values: $\vec{F} = (-1.6 \times 10^{-19}) (-200 \, \hat{j}) = 3.2 \times 10^{-17} \, \text{N} \, \hat{j}$

Using Newton's second law, $\vec{F} = m \vec{a}$: $\vec{a} = \frac{\vec{F}}{m} = \frac{3.2 \times 10^{-17}}{9.1 \times 10^{-31}} \, \hat{j}$ $\vec{a} = 3.52 \times 10^{13} \, \text{m/s}^2 \, \hat{j}$

Answer (a): The acceleration of the electron is: $\vec{a} = 3.52 \times 10^{13} \, \text{m/s}^2 \, \hat{j}$

Part (b) Time to travel through the region

The electron travels a distance $L = 0.1 \, \text{m}$ in the $x$-direction with constant velocity $v_0$: $t = \frac{L}{v_0}$ Substitute the values: $t = \frac{0.1}{3 \times 10^6} = 3.33 \times 10^{-8} \, \text{s}$

Answer (b): The time taken is: $t = 3.33 \times 10^{-8} \, \text{s}$

Part (c) Vertical displacement of the electron

The vertical displacement is caused by the acceleration $\vec{a}_y$ over time $t$. Using the kinematic equation: $y = \frac{1}{2} a_y t^2$ Substitute $a_y = 3.52 \times 10^{13} \, \text{m/s}^2$ and $t = 3.33 \times 10^{-8} \, \text{s}$: $y = \frac{1}{2} (3.52 \times 10^{13}) (3.33 \times 10^{-8})^2$ $y = 0.0195 \, \text{m} \, \approx 1.95 \, \text{cm}$

Answer (c): The vertical displacement is: $y = 1.95 \, \text{cm}$

Final Results:

• (a) Acceleration: $\vec{a} = 3.52 \times 10^{13} \, \text{m/s}^2 \, \hat{j}$.
• (b) Time: $t = 3.33 \times 10^{-8} \, \text{s}$.
• (c) Vertical displacement: $y = 1.95 \, \text{cm}$.

Do you want further details or clarification on any part?

Related Questions:

1. How would the results change if the field strength were doubled?
2. What is the work done by the field on the electron while in the field region?
3. How does the trajectory of the electron look in the field?
4. What would happen if the particle were a proton instead of an electron?
5. How can the vertical displacement be minimized in this setup?

Tip:

When solving problems involving charged particles in electric fields, always separate motion into perpendicular components for clarity.