Practice question
Question
Two charges \( 10 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 20 cm apart. What is the
potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2
\text{C}^{-2} \)).
Explanation
**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (10 × 10⁻⁶ × (-5 × 10⁻⁶)/0.2) = 9 × 10⁹ × (-50 × 10⁻¹²/0.2) = -2.25 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential
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