To analyze the situation, we can use Coulomb's law, which states that the electrical force between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
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 them.
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Original Situation:
- Charges: \( q_1 = +3 , \mu\text{C} \) and \( q_2 = -5 , \mu\text{C} \)
- Distance: \( r = 2 , \text{mm} = 0.002 , \text{m} \)
- Force (magnitude): \[ F_{original} = k \frac{|(+3 , \mu\text{C})(-5 , \mu\text{C})|}{(0.002)^2} \]
The force will be attractive because one charge is positive and the other is negative.
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New Situation:
- Charges: \( q_1 = +3 , \mu\text{C} \) and \( q_2 = +5 , \mu\text{C} \)
- Distance: \( r = 2 , \text{mm} = 0.002 , \text{m} \)
- Force (magnitude): \[ F_{new} = k \frac{|(+3 , \mu\text{C})(+5 , \mu\text{C})|}{(0.002)^2} \]
The force will be repulsive because both charges are positive.
Now, we can compare the magnitudes of the forces:
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Original force: \[ F_{original} = k \frac{15 , \mu\text{C}^2}{(0.002)^2} \]
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New force: \[ F_{new} = k \frac{15 , \mu\text{C}^2}{(0.002)^2} \]
Both forces have the same magnitude, specifically \( k \frac{15 , \mu\text{C}^2}{(0.002)^2} \), but they are in opposite directions (attractive for original, repulsive for new).
Conclusion: The correct answer is: same force, but in the opposite direction.