A dielectric slab of dielectric constant 3 is inserted into an uncharged capacitor `C_(1)`. The slab covers the entire volume of the capacitor now, this capacitor and an identical uncharged capacitor `C_(2)`, with air between the plates are placed in series and connected to a battery. after the capacitor are fully charged, the ratio of the electric field inside them, `(E_(1))/(E_(@))` is:

Options

A. (a)`(1)/(3)`
B. (b)`1`
C. (c)3
D. (d)9

Explanation

To solve the problem, we need to find the ratio of the electric fields inside two capacitors, one with a dielectric slab and the other with air between the plates, after they are connected in series to a battery.

### Step-by-Step Solution:

1. **Identify the Capacitors**:
- Let \( C_1 \) be the capacitor with a dielectric slab of dielectric constant \( K = 3 \).
- Let \( C_2 \) be the identical capacitor with air between the plates.

2. **Capacitance Calculation**:
- The capacitance of \( C_1 \) with the dielectric is given by:
\[ C_1 = K \cdot C = 3C \]
- The capacitance of \( C_2 \) (with air) remains:
\[ C_2 = C \]

3. **Total Capacitance in Series**:
- The total capacitance \( C_{total} \) for capacitors in series is given by:
\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Substituting the values:
\[ \frac{1}{C_{total}} = \frac{1}{3C} + \frac{1}{C} = \frac{1 + 3}{3C} = \frac{4}{3C} \]
- Therefore, the total capacitance is:
\[ C_{total} = \frac{3C}{4} \]

4. **Voltage Across Each Capacitor**:
- Let the voltage across the series combination be \( V \).
- The voltage across \( C_1 \) (with dielectric) is:
\[ V_1 = \frac{Q}{C_1} = \frac{Q}{3C} \]
- The voltage across \( C_2 \) (with air) is:
\[ V_2 = \frac{Q}{C_2} = \frac{Q}{C} \]

5. **Relation Between Voltages**:
- Since the total voltage \( V = V_1 + V_2 \):
\[ V = \frac{Q}{3C} + \frac{Q}{C} \]
- To combine these, find a common denominator:
\[ V = \frac{Q}{3C} + \frac{3Q}{3C} = \frac{4Q}{3C} \]

6. **Electric Field Calculation**:
- The electric field \( E \) in a capacitor is given by:
\[ E = \frac{V}{d} \]
- For \( C_1 \):
\[ E_1 = \frac{V_1}{d} = \frac{\frac{Q}{3C}}{d} = \frac{Q}{3Cd} \]
- For \( C_2 \):
\[ E_2 = \frac{V_2}{d} = \frac{\frac{Q}{C}}{d} = \frac{Q}{Cd} \]

7. **Finding the Ratio**:
- The ratio of the electric fields is:
\[ \frac{E_1}{E_2} = \frac{\frac{Q}{3Cd}}{\frac{Q}{Cd}} = \frac{1}{3} \]

### Final Answer:
The ratio of the electric field inside the two capacitors is:
\[ \frac{E_1}{E_2} = \frac{1}{3} \]