To determine which object has the strongest attraction to the object with a charge of -3 Coulombs, we can use Coulomb's Law, which states that the force \( F \) between two point charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The formula is:
\[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]
where
- \( F \) is the force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
- \( r \) is the distance between the charges.
We are interested in attraction, which occurs when one charge is positive and the other is negative.
Let's analyze each option:
-
An object with a -3 charge at a distance of 5 cm
- \( |q_1| = 3 \) C, \( |q_2| = 3 \) C, \( r = 0.05 \) m
- \( F = k \cdot \frac{3 \cdot 3}{(0.05)^2} \)
-
An object with a +3 charge at a distance of 5 cm
- \( |q_1| = 3 \) C, \( |q_2| = 3 \) C, \( r = 0.05 \) m
- \( F = k \cdot \frac{3 \cdot 3}{(0.05)^2} \)
- This is attractive.
-
An object with a -4 charge at a distance of 2 cm
- \( |q_1| = 3 \) C, \( |q_2| = 4 \) C, \( r = 0.02 \) m
- \( F = k \cdot \frac{3 \cdot 4}{(0.02)^2} \)
-
An object with a +3 charge at a distance of 2 cm
- \( |q_1| = 3 \) C, \( |q_2| = 3 \) C, \( r = 0.02 \) m
- \( F = k \cdot \frac{3 \cdot 3}{(0.02)^2} \)
- This is also attractive.
Now we will compare the magnitudes of the forces calculated for the attractive cases:
-
For the positive 3 charge at 2 cm: \[ F = k \cdot \frac{3 \cdot 3}{(0.02)^2} = k \cdot \frac{9}{0.0004} = k \cdot 22500 \]
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For the positive 3 charge at 5 cm: \[ F = k \cdot \frac{3 \cdot 3}{(0.05)^2} = k \cdot \frac{9}{0.0025} = k \cdot 3600 \]
-
For the negative 4 charge at 2 cm: \[ F = k \cdot \frac{3 \cdot 4}{(0.02)^2} = k \cdot \frac{12}{0.0004} = k \cdot 30000 \]
Thus, the strongest attraction comes from:
- An object with a -4 charge at a distance of 2 cm, yielding the highest force from our calculations.
Final Answer: An object with a -4 charge at a distance of 2 cm.
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