Final answer:
When capacitors are connected in parallel, their total capacitance is the sum of the individual capacitances. In this case, the total capacitance is 11.0 μF. The voltage across each capacitor is the total potential difference divided by the number of capacitors, which is 166.67 V.
Step-by-step explanation:
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, the three capacitors have capacitances of C₁ = 2.0 μF, C₂ = 3.0 μF, and C₃ = 6.0 μF, respectively. So the total capacitance is C = C₁ + C₂ + C₃. Plugging in the values, we get C = 2.0 μF + 3.0 μF + 6.0 μF = 11.0 μF.
Since the potential difference is applied across the combination, the voltage across each capacitor is the same. So the voltage across each capacitor is the total potential difference divided by the number of capacitors, which in this case is 500 V / 3 = 166.67 V.
The charge on each capacitor can be calculated using the formula Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage. So for each capacitor, we have Q₁ = C₁ × V = (2.0 μF) × (166.67 V) = 333.33 μC, Q₂ = C₂ × V = (3.0 μF) × (166.67 V) = 500 μC, and Q₃ = C₃ × V = (6.0 μF) × (166.67 V) = 1000 μC.