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. `(1)/(3)`
B. `1`
C. 3
D. 9

Explanation

To solve the problem, we need to analyze the behavior of the two capacitors when connected in series with a battery.

### Step 1: Understand the Capacitors
We have two capacitors:
- Capacitor \( C_1 \): Contains a dielectric slab with a dielectric constant \( K = 3 \).
- Capacitor \( C_2 \): Contains air (or vacuum) between its plates.

### Step 2: Electric Field in Capacitor \( C_1 \)
The electric field \( E_1 \) in a capacitor with a dielectric is given by the formula:
\[ E_1 = \frac{E_0}{K} \]
where \( E_0 \) is the electric field without the dielectric, and \( K \) is the dielectric constant.

Since \( K = 3 \), we have:
\[ E_1 = \frac{E_0}{3} \]

### Step 3: Electric Field in Capacitor \( C_2 \)
For capacitor \( C_2 \) which has air between the plates, the electric field \( E_2 \) is simply:
\[ E_2 = E_0 \]

### Step 4: Ratio of Electric Fields
Now, we need to find the ratio of the electric fields inside the two capacitors:
\[ \frac{E_1}{E_2} = \frac{\frac{E_0}{3}}{E_0} = \frac{1}{3} \]

### Final Answer
Thus, the ratio of the electric field inside capacitor \( C_1 \) to that in capacitor \( C_2 \) is:
\[ \frac{E_1}{E_2} = \frac{1}{3} \]

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