Two point charges q₁ = +8 μC and q₂ = +4 μC are 120 cm apart. Find Coulomb force magnitude and whether it attracts or repels.

Solution & explanation

1. Given: q₁ = 8 μC; q₂ = 4 μC; r = 120 cm; k = 9×10⁹ N m² C⁻².

2. Convert both charges to coulombs.

   $$q_1=\frac{8}{1000000}$$

3. The second charge is 4×10⁻⁶ C and separation is 1.20 m.

   $$r=\frac{120}{100}=1.2$$

4. Use Coulomb’s law for the magnitude.

   $$F=\frac{k\times q_1\times q_2}{r^2}$$

5. Substitute all SI values.

   $$F=\frac{288000000000}{1440000000000.0}$$

6. Calculate numerator and squared distance separately.

   $$F=\frac{0.288}{1.44}=0.2$$

7. State the result with units: F = 0.20 N, repulsive because both charges are positive.

F = 0.20 N, repulsive