To determine which object has the strongest attraction to an object with a charge of -3 Coulombs, we can use Coulomb's Law, which states that the force \( F \) between two charges is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
Where:
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the two charges,
- \( r \) is the distance between the centers of the two charges.
In this case, we're looking for an attractive force, which occurs between opposite charges (i.e., a negative charge and a positive charge).
Let's evaluate the options:
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An object with a -4 charge at a distance of 2 cm
- Charge: -4 C (Same sign, repulsive force)
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An object with a -3 charge at a distance of 5 cm
- Charge: -3 C (Same sign, repulsive force)
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An object with a +3 charge at a distance of 2 cm
- Charge: +3 C (Opposite sign, attractive force)
- Force: \( F = k \frac{|(-3)(+3)|}{(0.02)^2} = k \frac{9}{0.0004} = k \cdot 22500 \)
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An object with a +3 charge at a distance of 5 cm
- Charge: +3 C (Opposite sign, attractive force)
- Force: \( F = k \frac{|(-3)(+3)|}{(0.05)^2} = k \frac{9}{0.0025} = k \cdot 3600 \)
Comparing the attractive forces calculated:
- For the +3 charge at 2 cm: \( F = k \cdot 22500 \)
- For the +3 charge at 5 cm: \( F = k \cdot 3600 \)
The strongest attraction would be from the object with a +3 charge at a distance of 2 cm because it produces the greatest force due to being closer, despite the charge being equal to the object with a distance of 5 cm.
Final Answer: An object with a +3 charge at a distance of 2 cm.
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