Final answer:
The total stored energy after the capacitors are connected in parallel is 46.656 J.
Step-by-step explanation:
The total stored energy after the capacitors are connected in parallel can be found by adding up the energy stored in each individual capacitor. In this case, the capacitors have equal capacitances of 8.4 µF and 4.2 µF, and they were initially connected in series across a 36.0-V potential difference.
(a) To find the total energy stored in all three capacitors in the initial arrangement, we can use the formula for energy stored in a capacitor:
E = (1/2)CV²,
where E is the energy stored, C is the capacitance, and V is the potential difference. Plugging in the values for each capacitor, we get:
- For the first capacitor: E₁ = (1/2)(8.4 × 10⁻⁶ F)(36.0 V)² = 20.736 J
- For the second capacitor: E₂ = (1/2)(8.4 × 10⁻⁶ F)(36.0 V)² = 20.736 J
- For the third capacitor: E₃ = (1/2)(4.2 × 10⁻⁶ F)(36.0 V)² = 5.184 J
The total energy E_total stored in all three capacitors is the sum of the energy stored in each capacitor:
E_total = E₁ + E₂ + E₃ = 20.736 J + 20.736 J + 5.184 J = 46.656 J.