Question
Question
Suppose the electrostatic force between two electrons is F
. What is the electrostatic force between an electron and a proton that are the same distance apart?(1 point)
Responses
−F2
negative cap f over 2
−F
negative cap f
2F
2 cap f
F2
Suppose the electrostatic force between two electrons is F
. What is the electrostatic force between an electron and a proton that are the same distance apart?(1 point)
Responses
−F2
negative cap f over 2
−F
negative cap f
2F
2 cap f
F2
Answers
GPT-4o mini
The electrostatic force between two charged particles is given by Coulomb's law, which states that the force between two charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The charge of an electron (e) is approximately \(-1.6 \times 10^{-19}\) coulombs, and the charge of a proton (p) is approximately \(+1.6 \times 10^{-19}\) coulombs.
When considering the force between two electrons, the force is given as \(F\). The force between an electron and a proton is calculated as follows:
\[
F_{ep} = k \frac{|e| |p|}{r^2}
\]
Since \( |e| = |p| = 1.6 \times 10^{-19} \) C, we have:
\[
F_{ep} = k \frac{(1.6 \times 10^{-19})^2}{r^2}
\]
In comparison, the force between two electrons (\(F\)) can be written as:
\[
F_{ee} = k \frac{(1.6 \times 10^{-19})^2}{r^2}
\]
Noticing that the magnitudes for both forces depend on the same factors, we can conclude that the magnitudes are equal, but they will be of opposite signs since the electron and proton have opposite charges.
Thus, the force between the electron and proton (which is attractive) will be:
\[
F_{ep} = -F
\]
So, the electrostatic force between an electron and a proton that are the same distance apart is:
**−F** (negative cap f)
The charge of an electron (e) is approximately \(-1.6 \times 10^{-19}\) coulombs, and the charge of a proton (p) is approximately \(+1.6 \times 10^{-19}\) coulombs.
When considering the force between two electrons, the force is given as \(F\). The force between an electron and a proton is calculated as follows:
\[
F_{ep} = k \frac{|e| |p|}{r^2}
\]
Since \( |e| = |p| = 1.6 \times 10^{-19} \) C, we have:
\[
F_{ep} = k \frac{(1.6 \times 10^{-19})^2}{r^2}
\]
In comparison, the force between two electrons (\(F\)) can be written as:
\[
F_{ee} = k \frac{(1.6 \times 10^{-19})^2}{r^2}
\]
Noticing that the magnitudes for both forces depend on the same factors, we can conclude that the magnitudes are equal, but they will be of opposite signs since the electron and proton have opposite charges.
Thus, the force between the electron and proton (which is attractive) will be:
\[
F_{ep} = -F
\]
So, the electrostatic force between an electron and a proton that are the same distance apart is:
**−F** (negative cap f)