Question

The force between two small charged spheres having charges of \( 1 \times 10^{-7} \) C and \( 2 \times 10^{-7} \) C placed 20 cm apart in air is:

• A. \( 4.5 \times 10^{-2} \) N
• B. \( 4.5 \times 10^{-3} \) N
• C. \( 5.4 \times 10^{-2} \) N
• D. \( 5.4 \times 10^{-3} \) N

Hint:

For electrostatic force calculations using Coulomb’s Law: \[ F = \frac{k q_1 q_2}{r^2} \] - Convert all units to SI units (meters, Coulombs, Newtons). - Use \( k = 9 \times 10^9 \) N·m²/C² for air/vacuum.

Answer 1

The Correct Option is B

Step 1: Applying Coulomb’s Law. The electrostatic force between two charges is given by: \[ F = \frac{k q_1 q_2}{r^2} \] where: - \( k = 9 \times 10^9 \) N·m²/C² (Coulomb’s constant), - \( q_1 = 1 \times 10^{-7} \) C, - \( q_2 = 2 \times 10^{-7} \) C, - \( r = 20 \) cm \( = 0.2 \) m. 
Step 2: Substituting the values. \[ F = \frac{(9 \times 10^9) (1 \times 10^{-7}) (2 \times 10^{-7})}{(0.2)^2} \] \[ F = \frac{(9 \times 10^9) \times (2 \times 10^{-14})}{0.04} \] \[ F = \frac{18 \times 10^{-5}}{4 \times 10^{-2}} \] \[ F = 4.5 \times 10^{-3} { N} \] 
Final Answer: \[ \boxed{4.5 \times 10^{-3} { N}} \]