Two similar parallel plate capacitors each of capaciti `C_0` are connected in series The combination is connected with a voltage source of `V_0`. Now, seperation between the plates of one capacitor is increased by a distance `d` and the separation between the plates of another capacitor is decreased by the distance `d/2`. The distance between the plates of each capacitor was `d` before the chane in sepration. Then select the correct choice.

Options

A. the new capacity of the system will increase
B. the new capacity of the system will decrease
C. the new capacity of the system will remain same
D. data insufficient

Explanation

To solve the problem step by step, we will analyze the changes in capacitance for the two capacitors when their plate separations are altered.

### Step 1: Understand the initial conditions
Initially, we have two capacitors, each with capacitance \( C_0 \). The formula for the capacitance of a parallel plate capacitor is given by:

\[ C = \frac{A \epsilon_0}{d} \]

where \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the separation between the plates.

### Step 2: Calculate the initial capacitance
Since both capacitors are identical, we denote their capacitances as:

\[ C_1 = C_0 \quad \text{and} \quad C_2 = C_0 \]

When connected in series, the equivalent capacitance \( C_i \) of the two capacitors is given by:

\[ C_i = \frac{C_1 \cdot C_2}{C_1 + C_2} = \frac{C_0 \cdot C_0}{C_0 + C_0} = \frac{C_0^2}{2C_0} = \frac{C_0}{2} \]

### Step 3: Modify the capacitance due to changes in separation
Now, we modify the plate separations:
- For capacitor \( C_1 \), the separation is increased by \( d \), so the new separation becomes \( d + d = 2d \). The new capacitance \( C_1' \) is:

\[ C_1' = \frac{A \epsilon_0}{2d} = \frac{C_0}{2} \]

- For capacitor \( C_2 \), the separation is decreased by \( \frac{d}{2} \), so the new separation becomes \( d - \frac{d}{2} = \frac{d}{2} \). The new capacitance \( C_2' \) is:

\[ C_2' = \frac{A \epsilon_0}{\frac{d}{2}} = 2C_0 \]

### Step 4: Calculate the new equivalent capacitance
Now, we find the new equivalent capacitance \( C_i' \) of the modified capacitors:

\[ C_i' = \frac{C_1' \cdot C_2'}{C_1' + C_2'} = \frac{\left(\frac{C_0}{2}\right) \cdot (2C_0)}{\frac{C_0}{2} + 2C_0} \]

Calculating the numerator:

\[ \frac{C_0}{2} \cdot 2C_0 = C_0^2 \]

Calculating the denominator:

\[ \frac{C_0}{2} + 2C_0 = \frac{C_0}{2} + \frac{4C_0}{2} = \frac{5C_0}{2} \]

Thus, we have:

\[ C_i' = \frac{C_0^2}{\frac{5C_0}{2}} = \frac{2C_0}{5} \]

### Step 5: Compare the initial and new capacitances
Now we compare the initial capacitance \( C_i = \frac{C_0}{2} \) with the new capacitance \( C_i' = \frac{2C_0}{5} \).

To compare:

\[ C_i = \frac{C_0}{2} = \frac{5C_0}{10} \]
\[ C_i' = \frac{2C_0}{5} = \frac{4C_0}{10} \]

Since \( \frac{4C_0}{10} < \frac{5C_0}{10} \), we conclude that:

\[ C_i' < C_i \]

### Conclusion
The new capacity of the system will decrease.