Final answer:
Immediately after the switch is closed, the charge on a capacitor is zero. It will then charge up until it reaches its maximum charge, given by the capacitance times the voltage. The circuit reaches a steady state after a long time, with zero current through the resistors and the voltage across the capacitor equal to the voltage of the power supply.
Step-by-step explanation:
The charge on a capacitor immediately after the switch is closed in a circuit is zero. Since the capacitor starts uncharged, it will begin to accumulate charge as current flows through the circuit. This happens because a potential difference develops across the capacitor as it charges up. The current in the circuit immediately after the switch is closed is at its maximum because the voltage across the uncharged capacitor is initially zero, allowing maximum current to flow according to Ohm's law (I = V/R). As the capacitor charges, the voltage across it increases, which gradually reduces the current in the circuit until it eventually reaches zero when the capacitor is fully charged.
Addressing the question about the long-term behavior of the capacitor in the circuit, after a long time, the capacitor will be fully charged. At this point, the charge (Q) on the capacitor is given by Q = C × V, where C is the capacitance and V is the voltage. The current through each resistor a long time after the switch is closed will be zero, as the circuit reaches steady state and no current flows through the capacitor. The voltage across each resistor will also be zero for the same reason. The voltage across the capacitor will be equal to the voltage of the power supply, as it's fully charged. If the switch is then opened, the capacitor will start to discharge and the current will decay over time according to an exponential decay function. The time it takes for the current to drop to one fifth can be calculated using the RC time constant formula.