To determine the acceleration of an electron when it enters a region with a uniform electric field, we can follow these steps:
### Step-by-Step Solution:
1. Identify the Known Quantities:
- Charge of the electron, [tex]\( q \)[/tex]: [tex]\( 1.6 \times 10^{-19} \)[/tex] coulombs
- Mass of the electron, [tex]\( m \)[/tex]: [tex]\( 9.11 \times 10^{-31} \)[/tex] kilograms
- Electric field strength, [tex]\( E \)[/tex]: [tex]\( 200 \)[/tex] N/C (Newtons per Coulomb)
- Initial velocity, [tex]\( v_i \)[/tex]: [tex]\( 3 \times 10^6 \)[/tex] m/s (although not directly needed for acceleration)
- Horizontal length of the field region, [tex]\( d \)[/tex]: [tex]\( 0.1 \)[/tex] m (although not directly needed for acceleration)
2. Calculate the Force Acting on the Electron:
Using the formula for the electric force [tex]\( F \)[/tex] on a charge in an electric field:
[tex]\[
F = qE
\][/tex]
- Substituting the known values:
[tex]\[
F = (1.6 \times 10^{-19} \, \text{C}) \times (200 \, \text{N/C})
\][/tex]
- Simplifying the calculation:
[tex]\[
F = 3.2 \times 10^{-17} \, \text{N}
\][/tex]
3. Determine the Acceleration:
Using Newton's second law, which relates force, mass, and acceleration:
[tex]\[
F = ma \implies a = \frac{F}{m}
\][/tex]
- Substituting the values for [tex]\( F \)[/tex] and [tex]\( m \)[/tex]:
[tex]\[
a = \frac{3.2 \times 10^{-17} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}}
\][/tex]
- Simplifying the calculation:
[tex]\[
a \approx 3.512 \times 10^{13} \, \text{m/s}^2
\][/tex]
Hence, the acceleration of the electron in the electric field is [tex]\( 3.512 \times 10^{13} \)[/tex] meters per second squared.
