To determine which object has the strongest attraction to the charged object A (with a charge of -3 Coulombs), we can use Coulomb's law, which states that the force \( F \) between two charges \( q_1 \) and \( q_2 \) is given by the formula:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where:
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
- \( r \) is the distance between the charges.
Given that the most attractive force occurs between opposite charges, we identify that the object with a positive charge will attract the negative charge of object A.
Now, we will compare the provided options:
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An object with a charge of -4 at a distance of 2 cm:
- \( F = k \frac{|-3 \cdot -4|}{(0.02)^2} = k \frac{12}{0.0004} \) (repulsive, not attractive)
-
An object with a charge of +3 at a distance of 2 cm:
- \( F = k \frac{|-3 \cdot +3|}{(0.02)^2} = k \frac{9}{(0.02)^2} = k \frac{9}{0.0004} \)
-
An object with a charge of -3 at a distance of 5 cm:
- \( F = k \frac{|-3 \cdot -3|}{(0.05)^2} = k \frac{9}{(0.05)^2} = k \frac{9}{0.0025} \) (repulsive, not attractive)
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An object with a charge of +3 at a distance of 5 cm:
- \( F = k \frac{|-3 \cdot +3|}{(0.05)^2} = k \frac{9}{(0.05)^2} = k \frac{9}{0.0025} \)
Now we calculate the magnitudes of the attractive forces:
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For the +3 charge at 2 cm: \[ F = k \frac{9}{0.0004} = 22500 k \]
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For the +3 charge at 5 cm: \[ F = k \frac{9}{0.0025} = 3600 k \]
Comparing these values, the charge of +3 at a distance of 2 cm produces a much stronger attractive force (22500 k) compared to the charge of +3 at a distance of 5 cm (3600 k).
Thus, the object with the strongest attraction to the object with a charge of -3 Coulombs is:
An object with a +3 charge at a distance of 2 cm.