Question
if F1 and F2 are Forces acting on alpha particle and electron respectively in an electric field,then:
Answers
GPT-4o mini
In an electric field, the forces acting on charged particles can be determined using Coulomb's law and the formula for electric force. The force \( F \) on a charged particle in an electric field \( E \) is given by the equation:
\[
F = qE
\]
where \( q \) is the charge of the particle.
**For an alpha particle:**
An alpha particle consists of 2 protons and 2 neutrons, and has a charge of \( +2e \) (where \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) Coulombs). Thus, the force \( F_1 \) acting on the alpha particle in the electric field \( E \) can be expressed as:
\[
F_1 = (2e)E = 2(1.6 \times 10^{-19} \, \text{C})E
\]
**For an electron:**
An electron has a charge of \( -e \). Therefore, the force \( F_2 \) acting on the electron in the same electric field \( E \) is:
\[
F_2 = (-e)E = -(1.6 \times 10^{-19} \, \text{C})E
\]
**Comparison of the forces \( F_1 \) and \( F_2 \):**
- The force on the alpha particle \( F_1 \) is positive, indicating that it is attracted in the direction of the electric field (since it is positively charged).
- The force on the electron \( F_2 \) is negative, indicating that it is attracted in the opposite direction of the electric field (since it is negatively charged).
**Magnitude Comparison:**
The magnitudes of the forces can be compared as follows:
1. Magnitude of \( F_1 \):
\[
|F_1| = 2(1.6 \times 10^{-19} \, \text{C})|E|
\]
2. Magnitude of \( F_2 \):
\[
|F_2| = |(1.6 \times 10^{-19} \, \text{C})E| = (1.6 \times 10^{-19} \, \text{C})|E|
\]
Thus, the magnitude of the force on the alpha particle \( |F_1| \) is twice that of the force on the electron \( |F_2| \):
\[
|F_1| = 2 |F_2|
\]
In conclusion:
- Forces acting on the alpha particle and the electron are determined by their charges and the electric field.
- The alpha particle experiences a force that is twice as strong as that experienced by the electron, but in opposite directions due to their differing charges.
\[
F = qE
\]
where \( q \) is the charge of the particle.
**For an alpha particle:**
An alpha particle consists of 2 protons and 2 neutrons, and has a charge of \( +2e \) (where \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) Coulombs). Thus, the force \( F_1 \) acting on the alpha particle in the electric field \( E \) can be expressed as:
\[
F_1 = (2e)E = 2(1.6 \times 10^{-19} \, \text{C})E
\]
**For an electron:**
An electron has a charge of \( -e \). Therefore, the force \( F_2 \) acting on the electron in the same electric field \( E \) is:
\[
F_2 = (-e)E = -(1.6 \times 10^{-19} \, \text{C})E
\]
**Comparison of the forces \( F_1 \) and \( F_2 \):**
- The force on the alpha particle \( F_1 \) is positive, indicating that it is attracted in the direction of the electric field (since it is positively charged).
- The force on the electron \( F_2 \) is negative, indicating that it is attracted in the opposite direction of the electric field (since it is negatively charged).
**Magnitude Comparison:**
The magnitudes of the forces can be compared as follows:
1. Magnitude of \( F_1 \):
\[
|F_1| = 2(1.6 \times 10^{-19} \, \text{C})|E|
\]
2. Magnitude of \( F_2 \):
\[
|F_2| = |(1.6 \times 10^{-19} \, \text{C})E| = (1.6 \times 10^{-19} \, \text{C})|E|
\]
Thus, the magnitude of the force on the alpha particle \( |F_1| \) is twice that of the force on the electron \( |F_2| \):
\[
|F_1| = 2 |F_2|
\]
In conclusion:
- Forces acting on the alpha particle and the electron are determined by their charges and the electric field.
- The alpha particle experiences a force that is twice as strong as that experienced by the electron, but in opposite directions due to their differing charges.