Let's solve this problem step by step.

Part a) Capacitance of the Capacitor

The capacitance $C$ of a parallel plate capacitor is given by the formula:

$C = \frac{Q}{V}$

Where:

• $Q$ is the charge stored on the capacitor (3 × 10⁻⁶ C),
• $V$ is the voltage applied across the plates (100 V).

Substituting the given values:

$C = \frac{3 \times 10^{-6}}{100} = 3 \times 10^{-8} \, \text{F} = 30 \, \text{nF}$

So, the capacitance of the capacitor is $\boxed{30 \, \text{nF}}$.

Part b) Plate Area to Achieve this Capacitance

The capacitance of a parallel plate capacitor is also related to the area $A$ of the plates, the separation distance $d$ between them, and the permittivity of free space $\epsilon_0$ (since the material between the plates is a vacuum). The formula is:

$C = \frac{\epsilon_0 A}{d}$

Where:

• $\epsilon_0$ is the permittivity of free space, $8.85 \times 10^{-12} \, \text{F/m}$,
• $A$ is the area of each plate,
• $d$ is the separation between the plates (0.005 m).

We can rearrange this equation to solve for $A$:

$A = \frac{C \cdot d}{\epsilon_0}$

Substitute the known values:

$A = \frac{30 \times 10^{-9} \, \text{F} \times 0.005 \, \text{m}}{8.85 \times 10^{-12} \, \text{F/m}}$

$A = \frac{150 \times 10^{-12}}{8.85 \times 10^{-12}} \, \text{m}^2$

$A \approx 16.95 \, \text{m}^2$

So, the required plate area is approximately $\boxed{16.95 \, \text{m}^2}$.

---

Summary of Answers:

• Part a) The capacitance is $30 \, \text{nF}$.
• Part b) The plate area required is approximately $16.95 \, \text{m}^2$.

Let me know if you'd like any further details or explanations!

---

Follow-up Questions:

1. How does the permittivity of the material affect the capacitance of a parallel plate capacitor?
2. What would happen to the capacitance if the distance between the plates were doubled?
3. What are the effects of increasing the charge or voltage on the capacitance?
4. How would the capacitance change if a dielectric material was introduced between the plates?
5. Can you calculate the energy stored in the capacitor once it's fully charged?

Tip: When working with capacitors, always remember that the capacitance depends on the geometry of the plates, the distance between them, and the dielectric material present.