Question

an electron (mass ( m = 9.11 \times 10^{-31} , \text{kg} )) is accelerated from rest in the uniform field ( vec{e} ) (( e = 1.45 \times 10^4 , \text{n/c} )) between two thin parallel charged plates. the separation of the plates is ( 1.50 , \text{cm} ). the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, see the figure (figure 1)

figure

part b
show that the gravitational force can be ignored.
essay answers are limited to about 500 words (3800 characters maximum, including spaces).
3800 character(s) remaining

Explanation:

To show gravitational force can be ignored, compare electric force (\(F_E\)) and gravitational force (\(F_g\)) on the electron.

1. Electric Force: \(F_E = |q|E\), where \(q = 1.60\times10^{-19}\, \text{C}\) (electron charge), \(E = 1.45\times10^4\, \text{N/C}\).

\(F_E = (1.60\times10^{-19}\, \text{C})(1.45\times10^4\, \text{N/C}) = 2.32\times10^{-15}\, \text{N}\).

1. Gravitational Force: \(F_g = mg\), where \(m = 9.11\times10^{-31}\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\).

\(F_g = (9.11\times10^{-31}\, \text{kg})(9.8\, \text{m/s}^2) \approx 8.93\times10^{-30}\, \text{N}\).

1. Ratio: \(\frac{F_g}{F_E} \approx \frac{8.93\times10^{-30}}{2.32\times10^{-15}} \approx 3.85\times10^{-15}\), which is extremely small (negligible compared to \(F_E\)). Thus, gravitational force can be ignored.

Answer:

To determine if the gravitational force on the electron can be ignored, we compare it to the electric force acting on the electron:

1. Calculate the electric force (\(F_E\)):

The electric force on a charge \(q\) in an electric field \(E\) is \(F_E = |q|E\). For an electron, \(|q| = 1.60 \times 10^{-19}\, \text{C}\), and \(E = 1.45 \times 10^4\, \text{N/C}\).
\[
F_E = (1.60 \times 10^{-19}\, \text{C})(1.45 \times 10^4\, \text{N/C}) = 2.32 \times 10^{-15}\, \text{N}
\]

2. Calculate the gravitational force (\(F_g\)):

The gravitational force on a mass \(m\) is \(F_g = mg\), where \(m = 9.11 \times 10^{-31}\, \text{kg}\) (mass of electron) and \(g = 9.8\, \text{m/s}^2\) (acceleration due to gravity).
\[
F_g = (9.11 \times 10^{-31}\, \text{kg})(9.8\, \text{m/s}^2) \approx 8.93 \times 10^{-30}\, \text{N}
\]

3. Compare the forces:

The ratio of the gravitational force to the electric force is:
\[
\frac{F_g}{F_E} \approx \frac{8.93 \times 10^{-30}\, \text{N}}{2.32 \times 10^{-15}\, \text{N}} \approx 3.85 \times 10^{-15}
\]

This ratio is extremely small (on the order of \(10^{-15}\)), meaning \(F_g\) is negligible compared to \(F_E\). Thus, the gravitational force can be ignored when analyzing the electron’s motion in the electric field.