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MTH 102 PHY 102 Past Questions Objective Question
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Question

Determine the force between two protons placed at a distance of 1μm.
Options
A 2.3 &times; 10<sup>-13</sup> N, attraction
B 2.3 &times; 10<sup>-13</sup> N, repulsion
C 2.3 &times; 10<sup>-16</sup> N, attraction
D 2.3 &times; 10<sup>-16</sup> N, repulsion
Correct Answer
Correct Answer

Option D is the correct answer.

Detailed Explanation

The charge of a proton (q) is approximately 1.602 × 10-19 C. The distance (r) is 1 μm = 1 × 10-6 m. Coulomb's constant (k) is approximately 9 × 109 N·m2/C2.Coulomb's Law: F = k * |q1q2| / r2F = (9 × 109 N·m2/C2) * (1.602 × 10-19 C)2 / (1 × 10-6 m)2F = (9 × 109) * (2.5664 × 10-38) / (1 × 10-12)F = 23.0976 × 10(9 - 38 + 12) = 23.0976 × 10-17 NF ≈ 2.31 × 10-16 N.Since both protons have positive charges, the force between them is repulsive.

Hint

Use Coulomb's Law (F = kq1q2/r2) and remember the charge of a proton and the nature of the force between like charges.

Question Info

Course: MTH 102
Topic: PHY 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 545 of 930
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