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Two small, charged objects, Q1 and Q2, are some distance d apart from each other and there is a force F between them. What is the value of the force if Q1 is increased by a factor of two, Q2 is increased by a factor of 3, and d is increased by a factor of 5?

a) 0.24 F
b) 0.20 F
c) 1.2
d) 0.12F

User Artee
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1 Answer

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Final answer:

The new force between the two charged objects, given the increased charge and distance, is 0.24 times the original force F, according to Coulomb's Law.

Step-by-step explanation:

The force between two charged objects is described by Coulomb's Law, which states that the force (F) between two point charges is directly proportional to the product of their charges (Q1 and Q2) and inversely proportional to the square of the distance (d) between them. When Q1 is increased by a factor of 2 and Q2 is increased by a factor of 3, the product of their charges becomes 6 times larger because the increase is multiplicative (2 * 3). However, if d is increased by a factor of 5, the distance squared becomes 25 times greater. The force thus becomes 6/25 of the original force, since the factor increase of the charges is in the numerator and the factor increase of the distance squared is in the denominator of Coulomb's equation.

Therefore, the new force is 6/25 times the original force F, which can be simplified to 0.24 F. The correct answer is a) 0.24 F.

User SETI At Home
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