Question:medium

Two charges \( q_1 = 2 \, \mu \mathrm{C} \) and \( q_2 = -3 \, \mu \mathrm{C} \) are placed 10 cm apart in a vacuum. What is the magnitude and direction of the force between them?

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In Coulomb’s law, the force is attractive when the charges are of opposite signs and repulsive when the charges are of the same sign. Always convert the distance to meters in SI units for proper calculation.
Updated On: Nov 26, 2025
  • \( 5.4 \, \mathrm{N}, \, \text{Attractive} \)
  • \( 5.4 \, \mathrm{N}, \, \text{Repulsive} \)
  • \( 4.8 \, \mathrm{N}, \, \text{Attractive} \)
  • \( 4.8 \, \mathrm{N}, \, \text{Repulsive} \)
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The Correct Option is A

Solution and Explanation

To determine the electrostatic force between two point charges, Coulomb's Law is employed. This law states that the force \( F \) is calculated using the equation:\[F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|q_1 q_2|}{r^2},\]where: \( q_1 \) and \( q_2 \) represent the magnitudes of the charges, \( r \) is the separation distance between the charges, \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \mathrm{N \, m^2 \, C^{-2}} \) is Coulomb's constant.
Step 1: Input Values.Given values are:\[q_1 = 2 \, \mu \mathrm{C} = 2 \times 10^{-6} \, \mathrm{C}, \quad q_2 = -3 \, \mu \mathrm{C} = -3 \times 10^{-6} \, \mathrm{C}, \quad r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}.\]Substituting these into Coulomb's Law yields:\[F = 9 \times 10^9 \cdot \frac{|(2 \times 10^{-6})(-3 \times 10^{-6})|}{(0.1)^2}.\]
Step 2: Calculation.Calculate the product of the charges:\[|q_1 q_2| = |(2 \times 10^{-6})(-3 \times 10^{-6})| = 6 \times 10^{-12}.\]Substitute this back into the formula:\[F = 9 \times 10^9 \cdot \frac{6 \times 10^{-12}}{0.01}.\]Simplify the expression:\[F = 9 \cdot 6 \cdot 10^{-3} = 54 \times 10^{-3} = 5.4 \, \mathrm{N}.\]
Step 3: Direction Determination.As \( q_1 \) and \( q_2 \) have opposite signs (\( + \) and \( - \)), the force between them is attractive.
Result:The magnitude of the force is \( 5.4 \, \mathrm{N} \), and the force is attractive.
Therefore, the correct answer is \( \mathbf{(1)} \).
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