To determine which object would have the strongest attraction to an object with a charge of -3 Coulombs, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is given by the equation:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
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 charges, and
- \(r\) is the distance between the charges.
An attractive force occurs between opposite charges (negative and positive), while like charges (both negative or both positive) repel each other.
Let's analyze the options given:
A. An object with a -4 charge at a distance of 2cm:
- \(q_1 = -3 , C\), \(q_2 = -4 , C\), \(r = 0.02 , m\)
- Force is repulsive (same sign). So it won't attract.
B. An object with a -3 charge at a distance of 5cm:
- \(q_1 = -3 , C\), \(q_2 = -3 , C\), \(r = 0.05 , m\)
- Force is repulsive (same sign). So it won't attract.
C. An object with a +3 charge at a distance of 5cm:
- \(q_1 = -3 , C\), \(q_2 = +3 , C\), \(r = 0.05 , m\)
- Force is attractive.
- \(F = k \frac{|(-3)(3)|}{(0.05)^2} = k \frac{9}{0.0025} = k \cdot 3600\)
D. An object with a +3 charge at a distance of 2cm:
- \(q_1 = -3 , C\), \(q_2 = +3 , C\), \(r = 0.02 , m\)
- Force is attractive.
- \(F = k \frac{|(-3)(3)|}{(0.02)^2} = k \frac{9}{0.0004} = k \cdot 22500\)
Now we compare the forces:
- For option C (attraction at 5 cm): \(F \propto k \cdot 3600\)
- For option D (attraction at 2 cm): \(F \propto k \cdot 22500\)
Since \(22500\) is greater than \(3600\), object D (with +3 charge at 2 cm) will have the strongest attraction.
Therefore, the answer is:
D. An object with a +3 charge at a distance of 2cm.
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