Final answer:
After the 6.0-μF and 4.0-μF capacitors are connected, the total energy stored in the 6.0-μF capacitor at equilibrium is approximately 5.71 μJ.
Step-by-step explanation:
When two capacitors are connected as described, charge will flow between them until the voltage across both is the same. To find the final voltage (Vf) across the two capacitors when connected, we can use charge conservation. The total charge before the connection is the sum of the charges on each capacitor.
The initial charge on the 6.0-μF capacitor (Q1) is given by Q = C×V, so Q1 = 6.0 μF × 50 V = 300 μC. Similarly, the initial charge on the 4.0-μF capacitor (Q2) is Q2 = 4.0 μF × 34 V = 136 μC. The total initial charge (Qt) is Q1 + Q2 = 300 μC + 136 μC = 436 μC.
After connecting the capacitors together, the charge remains conserved, and the combined capacitance (Ct) is the sum of the individual capacitances, Ct = 6.0 μF + 4.0 μF = 10.0 μF. Using the formula Q = C×V, we can find Vf by rearranging to V = Q/C. Therefore, Vf = Qt / Ct = 436 μC / 10.0 μF = 43.6 V.
To find the final energy stored in the 6.0-μF capacitor, we can use the energy formula for a capacitor, E = 1/2 C V^2. The final energy (Ef) is Ef = 1/2 × 6.0 μF × (43.6 V)^2.
Calculating Ef yields Ef = 0.5 × 6.0 × 10^-6 F × 1902.96 V^2 ≈ 5.71 μJ. So the total energy stored in the 6.0-μF capacitor at equilibrium is approximately 5.71 μJ.