1. Concept: High voltages applied across standard capacitors result in significant electrical potential energy accumulation, proportional to the square of the voltage.
2. Formula: $$ U = \frac{1}{2} C V^2 $$
3. Solution: - Identify Capacitance: \( C = 1 \mu\text{F} = 1 \times 10^{-6} \text{ F} \).
- Identify Voltage: \( V = 4000 \text{ V} = 4 \times 10^3 \text{ V} \).
- Square the voltage: \( V^2 = (4 \times 10^3)^2 = 16 \times 10^6 \).
- Substitute: \( U = \frac{1}{2} (1 \times 10^{-6}) (16 \times 10^6) \).
- The positive and negative exponents cancel perfectly (\( 10^{-6} \times 10^6 = 1 \)).
- Calculate final value: \( U = \frac{1}{2} \times 16 = 8 \text{ Joules} \).
4. Why other options are incorrect: Option A forgets the \( 1/2 \) coefficient. Options B and C arise from failure to square the voltage correctly or from mishandling the scientific notation prefixes.
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