The force between two charged objects is determined by both the amount of charge on each object and the distance between them, according to Coulomb's Law. The formula for the electrostatic force \( F \) is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where:
- \( F \) is the magnitude of the force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the charges of the two objects, and
- \( r \) is the distance between the charges.
Given an object with a charge of -3 Coulombs (object A), the attraction will be strongest when:
- The other object has a positive charge (since opposite charges attract).
- The distance between the charges is minimized (to maximize the force).
Let's evaluate the options:
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An object with a -4 charge at a distance of 2 cm: This would repulse, not attract. So, it won't contribute to attraction.
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An object with a +3 charge at a distance of 5 cm: The force would be: \[ F = k \frac{|-3 \cdot 3|}{(5 , \text{cm})^2} \]
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An object with a +3 charge at a distance of 2 cm: The force would be: \[ F = k \frac{|-3 \cdot 3|}{(2 , \text{cm})^2} \]
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An object with a -3 charge at a distance of 5 cm: This would also repulse, not attract.
Now comparing the forces:
- The first option is repulsion.
- The second option gives a force proportional to \( \frac{9}{25} \).
- The third option gives a force proportional to \( \frac{9}{4} \).
- The fourth option is also repulsion.
Since \( \frac{9}{4} \) (attraction with the +3 charge at 2 cm) is greater than \( \frac{9}{25} \) (attraction with the +3 charge at 5 cm), the strongest attraction will occur with:
An object with a +3 charge at a distance of 2 cm.
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