An alpha particle contains two protons and two neutrons and has a net charge of $\mathbf{+}\mathbf{2}\mathit{e}$. The alpha particle is 1 mm away from a single proton, which has a charge of $\mathbf{-}\mathit{e}$. Which statement about the magnitudes of the electric forces between the particles is correct? (a) The force on the proton by the alpha particle is larger than the force on the alpha particle by the proton. (b) The force on the alpha particle by the proton is larger than the force on the proton by the alpha particle. (c) The forces are equal in magnitude. (d) Not enough information is given.

The force of attraction between two charged particles kept in the vicinity is called the Coulombic force of attraction. The magnitude of the force depends on the product of the two charges and inversly proporional to the distance between them. The Coulombic force of attraction is given by

$\mathit{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, r is the distance between the two charged particles, and q₁and q₂ are the charges.

It’s been given that the alpha particle contains two protons and two neutrons; hence it has a net charge of +2e.

It is also given that the single proton has a net charge of -e.

The distance between the alpha particle and the proton is 1 mm.

From Coulomb's law, we know that

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Substitute the values and solve as below.

$$\begin{gathered} F=9\times {10}^9\times \frac{2e\times e}{r^2} \\ =9\times {10}^9\times \frac{2{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1\times {10}^{-3}\right)}^2} \\ =4.608\times {10}^{-22}\text{N} \end{gathered}$$

It’s been given that the alpha particle contains two protons and two neutrons; hence it has a net charge of +2e.

It is also given that the single proton has a net charge of -e.

The distance between the alpha particle and the proton is 1 mm.

From Coulomb's law, we know that

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Substitute the values and solve as below.

$$\begin{gathered} F=9\times {10}^9\times \frac{2e\times e}{r^2} \\ =9\times {10}^9\times \frac{2{\left(1.6\times {10}^{-19}\right)}^2}{{\left(1\times {10}^{-3}\right)}^2} \\ =4.608\times {10}^{-22}\text{N} \end{gathered}$$

Therefore, the forces are equal in magnitude. The correct option is (c): The forces are equal in magnitude.