A \( 4 \, \mu\text{F} \) capacitor is charged to \( 250 \, \text{V} \). What is the energy stored in it?
**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. U = (1/2) C V² = (1/2) × 4 × 10⁻⁶ × (250)² . U = (1/2) × 4 × 10⁻⁶ × 62500 = 0.125 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
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density