To determine which object would have the strongest attraction to an object with a charge of -3 Coulombs, we can use Coulomb's law, which states that the force of attraction or repulsion between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's Law formula is:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where:
- \( F \) is the force between the charges,
- \( k \) is the Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the charges of the two objects,
- \( r \) is the distance between the charges.
Given the options:
-
An object with a -3 charge at a distance of 5 cm \[ F_1 = k \frac{|-3 \cdot -3|}{(0.05)^2} = k \frac{9}{0.0025} \] (repulsion, since both charges are negative)
-
An object with a -4 charge at a distance of 2 cm \[ F_2 = k \frac{|-3 \cdot -4|}{(0.02)^2} = k \frac{12}{0.0004} \] (repulsion, since both charges are negative)
-
An object with a +3 charge at a distance of 2 cm \[ F_3 = k \frac{|-3 \cdot 3|}{(0.02)^2} = k \frac{9}{0.0004} \] (attraction, since charges are opposite)
-
An object with a +3 charge at a distance of 5 cm \[ F_4 = k \frac{|-3 \cdot 3|}{(0.05)^2} = k \frac{9}{0.0025} \] (attraction, since charges are opposite)
Now we can calculate the forces for the attractive cases (F3 and F4):
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For F3: \[ F_3 = k \cdot \frac{9}{0.0004} = k \cdot 22500 \]
-
For F4: \[ F_4 = k \cdot \frac{9}{0.0025} = k \cdot 3600 \]
Since F3 (an object with a +3 charge at a distance of 2 cm) results in a higher force of attraction than F4, we conclude:
The object with the +3 charge at a distance of 2 cm would have the strongest attraction to the object with a charge of -3 Coulombs.