Question

The protons in a nucleus are approximately 2 ✕ 10−15 m apart. Consider the case where the protons are a distance d = 2.04 ✕ 10−15 m apart. Calculate the magnitude of the electric force (in N) between two protons at this distance?

Answer 1

The magnitude of the electric force between two protons at a distance of
`\( d = 2.04 * 10^(-15)` \) m is approximately
`\( 8.23 * 10^(-10) \) N.`

Step-by-step explanation:

In classical electrodynamics, the electric force between two point charges is given by Coulomb's Law:

`\[ F = (k \cdot |q_1 \cdot q_2|)/(r^2) \]`

Here,
`\( k \)` is Coulomb's constant, approximately equal to \( 8.99 \times
`10^9 \, \text{N m}^2/\text{C}^2 \)`. For protons, the charge (\( q \)) is the elementary charge
`(\( e \))`, which is approximately
`\( 1.6 * 10^(-19) \, \text{C} \)`.

Given
`\( d = 2.04 * 10^(-15) \)`m as the separation between protons, we substitute these values into the formula:

`\[ F = ((8.99 * 10^9) \cdot (1.6 * 10^(-19))^2)/((2.04 * 10^(-15))^2) \]`

Solving this expression yields the final answer of
`\( F \approx 8.23 * 10^(-10) \, \text{N} \)`. This value represents the magnitude of the electric force between the two protons at the specified distance.

In this calculation, it's essential to use consistent units and the correct values for physical constants to ensure accurate results. The formula and numerical substitutions provide a clear understanding of the physics involved in determining the electric force between charged particles.