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Question: Two charged particles are placed at a distance 1.0 cm apart. What is the minimum possible magnitude ...

Two charged particles are placed at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Answer

2.3 × 10⁻²⁴ N

Explanation

Solution

Given two charged particles separated by a distance r=1.0 cm=0.01 mr = 1.0\,\text{cm} = 0.01\,\text{m}, the electric force between them is given by Coulomb’s law:

F=kq1q2r2F = k \frac{q_1q_2}{r^2}.

Assuming the charges are the smallest possible nonzero charges, we take them to be elementary charges: q1=q2=e=1.602×10−19 Cq_1 = q_2 = e = 1.602 \times 10^{-19}\,\text{C}.

Substitute the values: F=(8.99×109) (1.602×10−19)2(0.01)2F = (8.99 \times 10^9)\,\frac{(1.602 \times 10^{-19})^2}{(0.01)^2}.

Calculate the numerator: (1.602×10−19)2≈2.566×10−38 C2(1.602 \times 10^{-19})^2 \approx 2.566 \times 10^{-38}\,\text{C}^2. (0.01)2=1.0×10−4 m2(0.01)^2 = 1.0 \times 10^{-4}\,\text{m}^2.

Thus, F≈8.99×109×2.566×10−381.0×10−4=8.99×109×2.566×10−34F \approx 8.99 \times 10^9 \times \frac{2.566 \times 10^{-38}}{1.0 \times 10^{-4}} = 8.99 \times 10^9 \times 2.566 \times 10^{-34}.

Multiply the numbers: 8.99×2.566≈23.068.99 \times 2.566 \approx 23.06, so, F≈23.06×10−25 N≈2.3×10−24 NF \approx 23.06 \times 10^{-25}\,\text{N} \approx 2.3 \times 10^{-24}\,\text{N}.

Thus, the minimum possible magnitude of the electric force acting on each charge is approximately 2.3×10−24 N2.3 \times 10^{-24}\,\text{N}.