Question
Download Solution PDFA charge Q1 of 3.0 μC is placed at point (36 cm, 0) and another charge Q2 of 9.0 μC is placed at a point (0, -27 cm).
Let i and j be the unit vectors in x- and y- directions, respectively. The force exerted by Q2 on Q1 is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Coulomb's Law:
- According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges
- And inversely proportional to the square of the distance between them.
- It acts along the line joining the two charges considered to be point charges.
Coulomb's Formula
F =
F =
In vector form,
where ϵ0 = permittivity in the air,
r = magnitude of the distance between two charge
Explanation:
Given,
The charge at point (36cm, 0), q1 = 3.0 μC
The charge at point (0,-27 cm), q2 = 9.0 μC
The position vector,
The magnitude of the position vector,
From Coulomb's law,
F = (0.96 N) i + (0.72 N) j
Last updated on Jun 19, 2025
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