The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit.
Two identical air capacitors are connected in series to a battery. Let Q be the charge supplied by the battery in this case. Now the space between the plates of one of the capacitors is filled with a dielectric slab of dielectric constant 3, while the battery remains connected. Let `DeltaQ` be the additional charge supplied by the battery after the insertion of slab. Calcualate `Q/DeltaQ`.

Explanation

To solve the problem, we need to calculate the ratio \( \frac{Q}{\Delta Q} \) where \( Q \) is the charge supplied by the battery when two identical air capacitors are connected in series, and \( \Delta Q \) is the additional charge supplied after inserting a dielectric slab in one of the capacitors.

### Step-by-Step Solution:

1. **Identify the capacitance of the capacitors**:
Let the capacitance of each air capacitor be \( C \).

2. **Calculate the equivalent capacitance of two capacitors in series**:
For two capacitors in series, the equivalent capacitance \( C_{eq} \) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \]
Therefore,
\[ C_{eq} = \frac{C}{2} \]

3. **Calculate the charge \( Q \) supplied by the battery**:
The charge \( Q \) supplied by the battery is given by:
\[ Q = C_{eq} \cdot V = \frac{C}{2} \cdot V \]

4. **Introduce the dielectric slab**:
When a dielectric slab with a dielectric constant \( K = 3 \) is inserted into one of the capacitors, the capacitance of that capacitor becomes:
\[ C_1' = K \cdot C = 3C \]
The other capacitor remains unchanged:
\[ C_2 = C \]

5. **Calculate the new equivalent capacitance \( C_{eq}' \)**:
The new equivalent capacitance with the dielectric slab is:
\[ \frac{1}{C_{eq}'} = \frac{1}{C_1'} + \frac{1}{C_2} = \frac{1}{3C} + \frac{1}{C} \]
Finding a common denominator:
\[ \frac{1}{C_{eq}'} = \frac{1}{3C} + \frac{3}{3C} = \frac{4}{3C} \]
Thus,
\[ C_{eq}' = \frac{3C}{4} \]

6. **Calculate the new charge \( Q' \) supplied by the battery**:
The new charge \( Q' \) supplied by the battery after the dielectric is inserted is:
\[ Q' = C_{eq}' \cdot V = \frac{3C}{4} \cdot V \]

7. **Calculate the additional charge \( \Delta Q \)**:
The additional charge \( \Delta Q \) is given by:
\[ \Delta Q = Q' - Q = \left(\frac{3C}{4} \cdot V\right) - \left(\frac{C}{2} \cdot V\right) \]
To simplify:
\[ \Delta Q = \frac{3CV}{4} - \frac{2CV}{4} = \frac{CV}{4} \]

8. **Calculate the ratio \( \frac{Q}{\Delta Q} \)**:
Now we can find the ratio:
\[ \frac{Q}{\Delta Q} = \frac{\frac{CV}{2}}{\frac{CV}{4}} = \frac{CV}{2} \cdot \frac{4}{CV} = 2 \]

### Final Answer:
\[ \frac{Q}{\Delta Q} = 2 \]