Two point charges q₁ = +3 μC and q₂ = -4 μC are located at position vectors r₁ = (2î+3ĵ+3k̂) m and r₂ = (î+ĵ+k̂) m, respectively. Taking 1 ÷ (4πε₀) = 9×10⁹ SI units, what is the force on q₂?
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Question

Two point charges q₁ = +3 μC and q₂ = -4 μC are located at position vectors r₁ = (2î+3ĵ+3k̂) m and r₂ = (î+ĵ+k̂) m, respectively. Taking 1 ÷ (4πε₀) = 9×10⁹ SI units, what is the force on q₂?

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G1 Physics · Accepted Answer

The vector from q₂ toward q₁ is r₁-r₂ = (2-1)î+(3-1)ĵ+(3-1)k̂ = î+2ĵ+2k̂. Its magnitude is r = √(1²+2²+2²) = 3 m. Because the charges have opposite signs, the force on q₂ is attractive and points from q₂ toward q₁. The force magnitude is F = k|q₁q₂| ÷ r². Substitute: F = 9×10⁹ × (3×10⁻⁶)(4×10⁻⁶) ÷ 3² = 12×10⁻³ N. The unit vector from q₂ toward q₁ is r̂ = (î+2ĵ+2k̂) ÷ 3. Therefore, F(vector) = 12×10⁻³ × (î+2ĵ+2k̂) ÷ 3 = 4×10⁻³(î+2ĵ+2k̂) = (4î+8ĵ+8k̂)×10⁻³ N. Correct option: B.