To determine which object would have the strongest attraction to the object with a charge of -3 Coulombs, we can apply Coulomb's Law, which states that the force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]
where:
- \( F \) is the electrostatic force,
- \( k \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the charges, and
- \( r \) is the distance between the charges.
Let's evaluate the options:
-
An object with a -3 charge at a distance of 5 cm:
- \( F = k \frac{|(-3) \cdot (-3)|}{(0.05)^2} = k \frac{9}{0.0025} \)
- This results in a repulsive force since both charges are negative.
-
An object with a +3 charge at a distance of 2 cm:
- \( F = k \frac{|(-3) \cdot (+3)|}{(0.02)^2} = k \frac{9}{0.0004} \)
- This is an attractive force.
-
An object with a -4 charge at a distance of 2 cm:
- \( F = k \frac{|(-3) \cdot (-4)|}{(0.02)^2} = k \frac{12}{0.0004} \)
- This results in a repulsive force as both charges are negative.
-
An object with a +3 charge at a distance of 5 cm:
- \( F = k \frac{|(-3) \cdot (+3)|}{(0.05)^2} = k \frac{9}{0.0025} \)
- This is also an attractive force, but at a greater distance than option 2.
Now, comparing the attractive forces:
- For the +3 charge at 2 cm: \( F = k \frac{9}{0.0004} \)
- For the +3 charge at 5 cm: \( F = k \frac{9}{0.0025} \)
The force at 2 cm is much greater than the force at 5 cm because the denominator is smaller for the 2 cm distance, meaning that \( \frac{9}{0.0004} > \frac{9}{0.0025} \).
Thus, the object with a +3 charge at a distance of 2 cm will have the strongest attraction to the object with a charge of -3 Coulombs.
43 answers