Practice question
Question
Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (-6, 0, 0) \) and \( (6, 0, 0)
\, \text{cm} \) in an external field \( E = 10^5/r^2 \, \text{N/C} \). What is the total potential
energy? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).
Explanation
**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Mutual energy: U₁₂ = 9 × 10⁹ × (8 × 10⁻⁶ × (-4 × 10⁻⁶)/0.12) = -2.4 J . External potential: V(r) = (10⁵/r) , at r = 0.06 m , V = (10⁵/0.06) = 1.67 × 10⁶ V . External energy: 8 × 10⁻⁶ × 1.67 × 10⁶ + (-4 × 10⁻⁶) ×
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