Question

The nucleus of helium atom contains two protons that are separated by a distance 3.0 $\times$ 10⁻¹⁵ m. The magnitude of the electrostatic force that each proton exerts on the other is

• A. 20.6 N
• B. 25.6 N
• C. 15.6 N
• D. 12.6 N

Hint:

When squaring numbers in scientific notation, remember to square the base number normally ($3^2 = 9$) and multiply the exponent power by 2 ($10^{-15 \times 2} = 10^{-30}$). Keeping your exponent tracks separate makes verification much easier!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The electrostatic interaction between two stationary point charges is governed by Coulomb's Law. Protons carry a known positive elementary charge, meaning they exert an equal and opposite repulsive electrostatic force on one another.

Step 2: Key Formula or Approach:
Coulomb's Law formula is written as: $$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$$ Where: $\frac{1}{4\pi\varepsilon_0} = k_e \approx 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$ $q_1 = q_2 = e \approx 1.6 \times 10^{-19}\text{ C}$ (the elementary charge of a proton) $r = 3.0 \times 10^{-15}\text{ m}$ (given separation distance)

Step 3: Detailed Explanation:
Substitute these physical constants and parameters into the formula: \[ F = (9 \times 10^9) \times \frac{(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(3.0 \times 10^{-15})^2} \] First, square the denominator in the fraction: \[ (3.0 \times 10^{-15})^2 = 9.0 \times 10^{-30} \] Now substitute this value back into our force equation: \[ F = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-30}} \] Cancel out the number 9 from both the numerator and denominator: \[ F = 2.56 \times \frac{10^9 \times 10^{-38}}{10^{-30}} \] Combine the powers of 10 using standard exponent rules: \[ F = 2.56 \times \frac{10^{-29}}{10^{-30}} = 2.56 \times 10^{-29 - (-30)} \] \[ F = 2.56 \times 10^1 = 25.6\text{ N} \]

Step 4: Final Answer:
The magnitude of the electrostatic force is 25.6 N.