To determine which of the given objects 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 \( q_1 \) and \( q_2 \) at a distance \( r \) is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where \( k \) is Coulomb's constant.
Let's evaluate the options:
-
An object with a +3 charge at a distance of 5 cm: \[ F = k \frac{|-3 \cdot 3|}{(0.05)^2} = k \frac{9}{0.0025} = k \cdot 3600 \]
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An object with a +3 charge at a distance of 2 cm: \[ F = k \frac{|-3 \cdot 3|}{(0.02)^2} = k \frac{9}{0.0004} = k \cdot 22500 \]
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An object with a -4 charge at a distance of 2 cm: \[ F = k \frac{|-3 \cdot -4|}{(0.02)^2} = k \frac{12}{0.0004} = k \cdot 30000 \]
-
An object with a -3 charge at a distance of 5 cm: \[ F = k \frac{|-3 \cdot -3|}{(0.05)^2} = k \frac{9}{0.0025} = k \cdot 3600 \]
Now we summarize the forces calculated:
- +3 charge, 5 cm: \( k \cdot 3600 \)
- +3 charge, 2 cm: \( k \cdot 22500 \)
- -4 charge, 2 cm: \( k \cdot 30000 \)
- -3 charge, 5 cm: \( k \cdot 3600 \)
The strongest attraction occurs for the third option: An object with a -4 charge at a distance of 2 cm which results in a force of \( k \cdot 30000 \).
Thus, the answer is: An object with a -4 charge at a distance of 2 cm.