Force Between Two Charges – Coulomb's Law Interactive Problem

Force Between Two Charges

Two charges, each of \(1\,\mu\text{C}\) (i.e. \(1\times10^{-6}\,\text{C}\)), are placed \(1\,\text{cm}\) (i.e. \(0.01\,\text{m}\)) apart in vacuum. Using Coulomb's law, determine the force between them.

Select the correct option:

• A) \(9 \times 10^{3}\,\text{N}\)
• B) \(90\,\text{N}\)
• C) \(1.1 \times 10^{-4}\,\text{N}\)
• D) \(10^{4}\,\text{N}\)

Detailed Explanation

According to Coulomb's law, the force \(F\) between two point charges is given by: \[ F = k \frac{q_1 q_2}{r^2}, \] where:

• \( k \) is Coulomb's constant \(\approx 9 \times 10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2\),
• \( q_1 = q_2 = 1\,\mu\text{C} = 1 \times 10^{-6}\,\text{C}\),
• \( r = 1\,\text{cm} = 0.01\,\text{m}\).

Substituting the values into Coulomb's law: \[ F = \frac{9 \times 10^{9} \times \left(1 \times 10^{-6}\right) \times \left(1 \times 10^{-6}\right)}{(0.01)^2}. \]

Compute the numerator: \[ 9 \times 10^{9} \times 1 \times 10^{-12} = 9 \times 10^{-3}. \]

Compute the denominator: \[ (0.01)^2 = 1 \times 10^{-4}. \]

Therefore, \[ F = \frac{9 \times 10^{-3}}{1 \times 10^{-4}} = 90\,\text{N}. \]

Hence, the correct answer is Option B.