Question

Two positive point charges are separated by a distance of 4 m in air. If the sum of the two charges is \( 36 \mu C \) and the electrostatic force between them is \( 0.18 \) N, then the bigger charge is:

• A. \( 30 \mu C \)
• B. \( 18 \mu C \)
• C. \( 20 \mu C \)
• D. \( 16 \mu C \)

Hint:

In Coulomb’s law problems, solving quadratic equations often helps determine individual charges when their sum is given.

Answer 1

The Correct Option is C

Using Coulomb’s law: \[ F = k \frac{q_1 q_2}{r^2} \] where: - \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \), - \( r = 4 \) m, - \( q_1 + q_2 = 36 \mu C \), - \( F = 0.18 \) N. Substituting values: \[ 0.18 = 9 \times 10^9 \times \frac{q_1 q_2}{4^2} \] \[ q_1 q_2 = \frac{0.18 \times 16}{9 \times 10^9} \times 10^{-12} \] \[ q_1 q_2 = 3.2 \times 10^{-12} \text{C}^2 \] Solving for \( q_1, q_2 \): \[ q_1, q_2 = \frac{36 \pm \sqrt{36^2 - 4 \times 3.2}}{2} \] \[ q_1 = 20 \mu C, \quad q_2 = 16 \mu C \] Thus, the larger charge is \( 20 \mu C \).