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Coulomb's Law

Easyphysics

Two identical point charges each of magnitude 6 (\mu C) are placed 30 cm apart in vacuum. What is the magnitude of the electrostatic force between them?

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About This Question

Subject
physics
Chapter
electrostatics
Topic
coulomb's law
Difficulty
Easy
Year
2025
Tags
Coulomb's lawpoint chargeselectrostatic forceinverse square lawvacuum permittivity

Solution

Correct Answer:

3.6 N

Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, given by (F = k\frac{q_1 q_2}{r^2}), where (k = 9 \times 10^9) N·m²/C². This is the fundamental law governing all electrostatic interactions between stationary charges in vacuum. Here, (q_1 = q_2 = 6 \times 10^{-6}) C and (r = 0.30) m. Substituting: (F = 9 \times 10^9 \times \frac{(6 \times 10^{-6})^2}{(0.30)^2} = 9 \times 10^9 \times \frac{36 \times 10^{-12}}{0.09} = 9 \times 10^9 \times 4 \times 10^{-10} = 3.6) N. Option 1.2 N is incorrect because it corresponds to tripling the denominator incorrectly. Option 2.4 N is incorrect because it arises from an error in the square of the charge magnitude. Option 4.8 N is incorrect because it results from using (r = 0.30) m without squaring it in the denominator. This directly tests the NCERT Coulomb's Law formula and the standard JEE approach of careful unit substitution. Plausibility check: the result of 3.6 N is reasonable for microcoulomb charges separated by 30 cm, consistent with everyday electrostatic scales.

This easy difficulty physics question is from the chapter electrostatics, covering the topic of coulomb's law. It appeared in the 2025 exam.

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