At some instant the velocity components of an electron moving between two charged parallel plates are `v_x=1.5xx10^5m//s` and `v_y=3.0xx10^6 m//s`.Suppose that the electric field between the plates is given by `E = (120 N//C) hatj`
. (a) What is the acceleration of the electron?
(b) What will be the velocity of the electron after its x-coordinate has changed by 2.0 cm?

Explanation

To solve the problem step by step, we will break it down into two parts as stated in the question.

### Part (a): Finding the acceleration of the electron

1. **Identify the charge of the electron**:
The charge of an electron (q) is given as:
\[ q = -1.6 \times 10^{-19} \text{ C} \]

2. **Identify the electric field**:
The electric field (E) between the plates is given as:
\[ E = 120 \hat{j} \text{ N/C} \]

3. **Calculate the force on the electron**:
The force (F) acting on the electron due to the electric field can be calculated using the formula:
\[ F = qE \]
Substituting the values:
\[ F = (-1.6 \times 10^{-19} \text{ C})(120 \hat{j} \text{ N/C}) = -1.92 \times 10^{-17} \hat{j} \text{ N} \]

4. **Calculate the mass of the electron**:
The mass (m) of the electron is:
\[ m = 9.1 \times 10^{-31} \text{ kg} \]

5. **Calculate the acceleration**:
The acceleration (a) can be calculated using Newton's second law:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{-1.92 \times 10^{-17} \hat{j}}{9.1 \times 10^{-31}} \approx -2.11 \times 10^{13} \hat{j} \text{ m/s}^2 \]

### Part (b): Finding the velocity of the electron after its x-coordinate has changed by 2.0 cm

1. **Convert the distance**:
The change in x-coordinate (Δx) is given as:
\[ \Delta x = 2.0 \text{ cm} = 2.0 \times 10^{-2} \text{ m} \]

2. **Calculate the time taken to change the x-coordinate**:
Since there is no acceleration in the x-direction, the time (t) can be calculated using:
\[ t = \frac{\Delta x}{v_x} \]
Substituting the values:
\[ v_x = 1.5 \times 10^{5} \text{ m/s} \]
\[ t = \frac{2.0 \times 10^{-2}}{1.5 \times 10^{5}} \approx 1.33 \times 10^{-7} \text{ s} \]

3. **Calculate the new velocity in the y-direction**:
The new velocity in the y-direction (v_y') can be calculated using:
\[ v_y' = v_y + a_y \cdot t \]
Where:
\[ v_y = 3.0 \times 10^{6} \text{ m/s} \]
\[ a_y = -2.11 \times 10^{13} \text{ m/s}^2 \]
Substituting the values:
\[ v_y' = 3.0 \times 10^{6} + (-2.11 \times 10^{13}) \cdot (1.33 \times 10^{-7}) \approx 2.0 \times 10^{5} \text{ m/s} \]

4. **Final velocity of the electron**:
The final velocity vector (v) of the electron can be expressed as:
\[ v = v_x \hat{i} + v_y' \hat{j} \]
Substituting the values:
\[ v = (1.5 \times 10^{5} \hat{i} + 2.0 \times 10^{5} \hat{j}) \text{ m/s} \]

### Summary of Results:
- **Acceleration of the electron**:
\[ a \approx -2.11 \times 10^{13} \hat{j} \text{ m/s}^2 \]
- **Velocity of the electron after changing x-coordinate by 2.0 cm**:
\[ v \approx (1.5 \times 10^{5} \hat{i} + 2.0 \times 10^{5} \hat{j}) \text{ m/s} \]