Question

The electrostatic force between a proton and an electron for certain distance of separation is F₁ and that between an electron and positron at the same distance of separation is F₂. Then the ratio F₁: F₂ is

• A. 1:1
• B. 1:2
• C. 1879:1
• D. 1:1879
• E. 2:1

Answer 1

The Correct Option is A

The electrostatic force between two charges is given by Coulomb's law: \[ F = k_e \frac{q_1 q_2}{r^2} \] where \( F \) is the electrostatic force, \( k_e \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation distance.
The charge of a proton is \( +e \) and that of an electron is \( -e \), where \( e \) is the elementary charge.
The charge of an electron is \( -e \) and that of a positron is \( +e \). In both cases, the charges involved have the same magnitude \( e \), and the distance \( r \) is the same.
Therefore, the electrostatic force \( F_1 \) between the proton and the electron is equal to the electrostatic force \( F_2 \) between the electron and the positron.
Hence, the ratio of the forces is: \[ F_1 : F_2 = 1 : 1 \]

The correct option is (A) : \(1:1\)

Answer 2

The electrostatic force between two charges is given by Coulomb’s law: 
 
$$ F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2} $$ 
Where:

• \( q_1, q_2 \) are the magnitudes of the two charges
• \( r \) is the distance between them

Step 1: Proton and electron
Charges: \( +e \) and \( -e \) 
So, $$ F_1 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} $$ 
Step 2: Electron and positron
Charges: \( -e \) and \( +e \) 
So, $$ F_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} $$ 
Hence, $$ \frac{F_1}{F_2} = \frac{e^2/r^2}{e^2/r^2} = 1 $$ 
Final Answer: \( \boxed{1:1} \)