A resistor, capacitor, switch, and ideal battery are in series. Originally the capacitor is uncharged. The switch is then closed, allowing current of flow
While the current is flowing, the potential difference across the capacitor is

Options

A. Increasing
B. Decreasing
C. Fixed
D. First increase then decreases

Answer

To solve the problem, we need to analyze the behavior of the capacitor in an RC (resistor-capacitor) circuit when the switch is closed. Let's break down the steps:

### Step 1: Understand the Circuit
We have a series circuit consisting of a resistor (R), a capacitor (C), a switch (S), and an ideal battery (V). Initially, the capacitor is uncharged.

**Hint:** Identify the components in the circuit and their arrangement.

### Step 2: Closing the Switch
When the switch is closed, the circuit is completed, allowing current to flow from the battery through the resistor and into the capacitor.

**Hint:** Remember that closing the switch allows current to start flowing in the circuit.

### Step 3: Current Flow and Charging of the Capacitor
As the current flows, the capacitor begins to charge. The charging of the capacitor causes the voltage across it to increase over time. The relationship between the current (I), voltage (V), and resistance (R) can be described by the equation:

\[ V_C(t) = V(1 - e^{-t/RC}) \]

Where:
- \( V_C(t) \) is the voltage across the capacitor at time \( t \),
- \( V \) is the voltage of the battery,
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( e \) is the base of the natural logarithm.

**Hint:** The voltage across the capacitor increases exponentially as it charges.

### Step 4: Potential Difference Across the Capacitor
At the moment the switch is closed, the potential difference across the capacitor starts at 0 volts (since it is initially uncharged) and increases towards the battery voltage \( V \) as the capacitor charges.

**Hint:** The potential difference across the capacitor is initially zero and increases over time.

### Step 5: Long-Term Behavior
Eventually, when the capacitor is fully charged, the potential difference across it will equal the voltage of the battery \( V \), and the current will cease to flow.

**Hint:** Consider what happens to the circuit after a long time has passed.

### Conclusion
While the current is flowing, the potential difference across the capacitor is increasing from 0 volts towards the battery voltage \( V \).

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