An air parallel plate capacitor has capacity C. When the area and distance between the plates is doubled, the capacitance is `C_(1)`, then `(C_(1))/(C )` is -

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze how the capacitance of a parallel plate capacitor changes when the area and distance between the plates are modified.

### Step-by-Step Solution:

1. **Understand the Formula for Capacitance**:
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \cdot \epsilon_0}{D} \]
where:
- \( C \) = capacitance
- \( A \) = area of the plates
- \( \epsilon_0 \) = permittivity of free space (a constant)
- \( D \) = distance between the plates

2. **Identify the Changes**:
According to the problem, both the area \( A \) and the distance \( D \) are doubled. Therefore:
- New area \( A_1 = 2A \)
- New distance \( D_1 = 2D \)

3. **Calculate the New Capacitance**:
Substitute the new values into the capacitance formula to find the new capacitance \( C_1 \):
\[ C_1 = \frac{A_1 \cdot \epsilon_0}{D_1} = \frac{(2A) \cdot \epsilon_0}{2D} \]

4. **Simplify the Expression**:
Simplifying the equation gives:
\[ C_1 = \frac{2A \cdot \epsilon_0}{2D} = \frac{A \cdot \epsilon_0}{D} = C \]

5. **Calculate the Ratio of the New Capacitance to the Original Capacitance**:
Now, we need to find the ratio \( \frac{C_1}{C} \):
\[ \frac{C_1}{C} = \frac{C}{C} = 1 \]

### Final Answer:
Thus, the ratio \( \frac{C_1}{C} \) is \( 1 \).

---