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To solve part **(c)** of the question, let's first break down the key details provided:

### Part c: Find the energy stored in the capacitor 0.5 seconds after the discharge has started.

**Known values:**
- Capacitance \( C = 2.2 \, \mu\text{F} = 2.2 \times 10^{-6} \, \text{F} \)
- Initial potential difference (pd) across the capacitor \( V_0 = 6.0 \, \text{V} \)
- Resistance \( R = 100 \, \text{k}\Omega = 100 \times 10^3 \, \Omega \)
- Time \( t = 0.5 \, \text{s} \)

### Step 1: Calculate the voltage across the capacitor at \( t = 0.5 \, \text{s} \)
The voltage across a discharging capacitor is given by the formula:
\[
V(t) = V_0 e^{-\frac{t}{RC}}
\]
Where:
- \( V_0 = 6.0 \, \text{V} \) (initial voltage),
- \( R = 100 \times 10^3 \, \Omega \),
- \( C = 2.2 \times 10^{-6} \, \text{F} \),
- \( t = 0.5 \, \text{s} \).

Let's calculate \( V(0.5) \).

### Step 2: Calculate the energy stored in the capacitor at \( t = 0.5 \, \text{s} \)
The energy stored in a capacitor is given by the formula:
\[
E = \frac{1}{2} C V(t)^2
\]
Once we have \( V(t) \), we can calculate the energy stored at that time.

Let's go through the calculations.After 0.5 seconds, the voltage across the capacitor is approximately \( 0.618 \, \text{V} \).

The energy stored in the capacitor at this time is approximately \( 4.20 \times 10^{-7} \, \text{J} \) or \( 0.420 \, \mu\text{J} \).

Do you need further clarification or details?

### Related questions:
1. How does the time constant \( \tau = RC \) affect the discharge rate of a capacitor?
2. What is the significance of the exponential decay in the voltage across a discharging capacitor?
3. How would the energy stored in the capacitor change if the resistance was doubled?
4. How long would it take for the capacitor to lose 90% of its energy?
5. What happens to the capacitor's stored energy after a very long time of discharging?

### Tip:
Always check the time constant \( \tau = RC \) to understand how quickly the voltage and energy decay in RC circuits.

A 2.2 μF capacitor is charged to a pd of 6.0V and then discharged through a 100 kΩ resistor. Calculate: the energy stored at this time.

Learn how to calculate the energy stored in a 2.2 μF capacitor discharged through a 100 kΩ resistor after 0.5 seconds. This detailed solution covers concepts such as capacitor discharge, exponential decay, and energy storage in capacitors, using key formulas.

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