Practice question
Question
A point charge \( Q = 5 \times 10^{-9} \, \text{C} \) is placed at the origin. What is the potential at
a point 10 m away from the charge? (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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (5 × 10⁻⁹/10) = 9 × 10⁹ × 0.5 × 10⁻⁹ = 4.5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC
Discussion
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