Given: Q1 = 3.0 × 10-5 CQ2 = 4.0 × 10-5 C
K = 9.00 × 109 N ×
r = 3.0 m, determine F.
_____ N
Answers
The force is obtained from the charges Q₁ = 3×10⁻⁵ C and Q₂ = 4×10⁻⁵ C and distance r = 3 m is 1.2 N.
What is Coloumb's law?
Coloumb's law is defined as the force of attraction and repulsion between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.
The attractive and repulsive force, F∝ (q₁×q₂) / r². It is also known as F = k (q₁×q₂) / r², where k is the proportional constant. The unit of force is newton N.
From the given,
Q₁ = 3×10⁻⁵ C
Q₂ = 4×10⁻⁵ C
r = 3m
k = 9×10⁹ Nm²C⁻²
The attractive and repulsive force,
F = k (q₁×q₂) / r²
= (9×10⁹× 3×10⁻⁵×4×10⁻⁵) / (3)²
= 12×10⁻¹
= 1.2 N
Force = 1.2 N
Thus, the force of repulsion and attraction of given charges is 1.2 N.
To know more about Coloumb's law:
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F=9 x 10^9 x 3 x 10^-5 x 4 x 10^-5 / 9
F=12 x 10^-1
F=1.2 N
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