Question

The ratio of the magnitudes of electrostatic force between an electron and a proton separated by a distance \( r \) to that between a proton and an alpha particle separated by the same distance \( r \) is:

• A. 1:1
• B. 1:4
• C. 4:1
• D. 2:1
• E. 1:2

Hint:

The electrostatic force between charges is directly proportional to the product of the charges.

Answer 1

Answer: (E)

Using Coulomb’s law, the force between two charges \( q_1 \) and \( q_2 \) is given by: \[ F = k \frac{q_1 q_2}{r^2} \] For an electron (\( e \)) and a proton (\( e \)), and a proton (\( e \)) and an alpha particle (\( 2e \)): \[ F_{e-p} = k \frac{e \cdot e}{r^2}, \quad F_{p-\alpha} = k \frac{e \cdot 2e}{r^2} = 2k \frac{e^2}{r^2} \] \[ {Ratio} = \frac{F_{e-p}}{F_{p-\alpha}} = \frac{k \frac{e^2}{r^2}}{2k \frac{e^2}{r^2}} = \frac{1}{2} \]