Answer:
a. Here is a free-body diagram depicting the forces acting on the electron:
```
F_Electric
↑
│
――――――┼――→ v_initial
Weight │
――――――↓
```
The weight of the electron acts vertically downwards, while the electric force (F_Electric) acts vertically upwards due to the electric field generated by the charged plates.
b. To calculate the vertical deflection, we can use the equations of motion. The electron experiences a vertical force equal to the product of its charge (q) and the electric field intensity (E), which can be written as F_Electric = qE.
Using the formula for the force acting on an object (F = ma), and considering that the only vertical force acting on the electron is the electric force, we have:
qE = ma
Since the mass of the electron (m) is known, we can rearrange the equation to solve for the acceleration (a):
a = qE/m
The vertical deflection can be calculated using the equations of motion for uniformly accelerated motion:
y = v_initial * t + (1/2) * a * t^2
Where:
- y is the vertical deflection
- v_initial is the initial vertical velocity (in this case, 0)
- t is the time taken to travel the horizontal distance (0.0345m)
- a is the acceleration
To calculate the time (t), we can use the equation:
s = v_initial * t + (1/2) * a * t^2
Where:
- s is the horizontal distance (0.0345m)
- v_initial is the initial vertical velocity (0)
- t is the time
Simplifying the equation, we have:
s = (1/2) * a * t^2
Rearranging and solving for t:
t^2 = (2s) / a
t = sqrt((2 * s) / a)
Plugging in the given values for the electric field intensity (E = 357 N/C), initial velocity (v_initial = 42000000 m/s), horizontal distance (s = 0.0345m), and charge of an electron (q = 1.6 x 10^-19 C) along with the mass of an electron (m = 9.11 x 10^-31 kg), we can determine the vertical deflection.
Please note that dividing by a tiny mass and a large velocity may yield extremely large values. Also, the given electric field intensity seems unusually high. Please double-check the values and equations provided.