A parallel plate capacitor is to be constructed which can store `q = 10 muC` charge at `V = 1000` volt. The minimum plate area of the capacitor is required to be `A_(1)` when space between the plates has air. If a dielectric of constant `K = 3` is used between the plates, the minimum plate area required to make such a capacitor is `A_(2)`. The breakdown field for the dielectric is 8 times that of air. Find `(A_(1))/(A_(2))`

Answer

To solve the problem, we need to find the ratio of the minimum plate areas \( A_1 \) and \( A_2 \) for a parallel plate capacitor with and without a dielectric.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a capacitor that stores a charge \( q = 10 \, \mu C = 10 \times 10^{-6} \, C \) at a voltage \( V = 1000 \, V \).
- We need to find the minimum plate area \( A_1 \) when the space between the plates has air and \( A_2 \) when a dielectric with constant \( K = 3 \) is used.

2. **Capacitance Formula**:
- The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon A}{d} \]
where \( \varepsilon \) is the permittivity of the material between the plates, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

3. **Charge and Capacitance Relationship**:
- The relationship between charge, capacitance, and voltage is given by:
\[ q = C \cdot V \]
- Rearranging gives:
\[ C = \frac{q}{V} \]

4. **Finding \( A_1 \) (Air)**:
- For air, the permittivity \( \varepsilon_0 \) is used:
\[ C_1 = \frac{\varepsilon_0 A_1}{d} \]
- Thus:
\[ A_1 = \frac{C_1 \cdot d}{\varepsilon_0} \]
- Substituting for \( C_1 \):
\[ A_1 = \frac{q \cdot d}{\varepsilon_0 V} \]

5. **Finding \( A_2 \) (With Dielectric)**:
- When a dielectric is present, the capacitance becomes:
\[ C_2 = \frac{K \varepsilon_0 A_2}{d} \]
- Thus:
\[ A_2 = \frac{C_2 \cdot d}{K \varepsilon_0} \]
- Substituting for \( C_2 \):
\[ A_2 = \frac{q \cdot d}{K \varepsilon_0 V} \]

6. **Finding the Ratio \( \frac{A_1}{A_2} \)**:
- Now we can find the ratio:
\[ \frac{A_1}{A_2} = \frac{\frac{q \cdot d}{\varepsilon_0 V}}{\frac{q \cdot d}{K \varepsilon_0 V}} = \frac{1}{\frac{1}{K}} = K \]
- Given \( K = 3 \):
\[ \frac{A_1}{A_2} = 3 \]

7. **Considering Breakdown Fields**:
- The problem states that the breakdown field for the dielectric is 8 times that of air. This means:
\[ \frac{E_{\text{dielectric}}}{E_{\text{air}}} = 8 \]
- The electric field \( E \) is related to voltage and distance:
\[ E = \frac{V}{d} \]
- Therefore, we can incorporate this into our calculations, but since we are only interested in the ratio of areas, we can conclude:
\[ \frac{A_1}{A_2} = K \cdot \frac{E_{\text{dielectric}}}{E_{\text{air}}} = 3 \cdot 8 = 24 \]

### Final Answer:
\[ \frac{A_1}{A_2} = 24 \]