To solve the problem of two identical parallel plate capacitors connected in parallel and in series, we will analyze each case step by step.
### Step 1: Define Initial Capacitance
Let the capacitance of each capacitor be \( C \). For parallel plate capacitors, the capacitance is given by:
\[
C = \frac{\varepsilon_0 A}{D}
\]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( D \) is the separation between the plates.
### Step 2: Case 1 - Capacitors in Parallel
When two capacitors are connected in parallel, the total capacitance \( C_{eq} \) is the sum of the individual capacitances:
\[
C_{eq} = C_1 + C_2
\]
Initially, both capacitors have the same capacitance \( C \), so:
\[
C_{eq, initial} = C + C = 2C
\]
### Step 3: Modify the Capacitor Distances
In this case, we bring the plates of one capacitor closer by a distance \( a \) and move the plates of the other capacitor apart by the same distance \( a \). Thus, the new separations are:
- For the first capacitor: \( D_1 = D - a \)
- For the second capacitor: \( D_2 = D + a \)
The new capacitances are:
\[
C_1 = \frac{\varepsilon_0 A}{D - a}
\]
\[
C_2 = \frac{\varepsilon_0 A}{D + a}
\]
### Step 4: Calculate New Equivalent Capacitance in Parallel
Now, the new equivalent capacitance in parallel is:
\[
C_{eq, new} = C_1 + C_2 = \frac{\varepsilon_0 A}{D - a} + \frac{\varepsilon_0 A}{D + a}
\]
Combining these fractions gives:
\[
C_{eq, new} = \varepsilon_0 A \left( \frac{D + a + D - a}{(D - a)(D + a)} \right) = \varepsilon_0 A \left( \frac{2D}{D^2 - a^2} \right)
\]
Thus, we find that:
\[
C_{eq, new} > C_{eq, initial} \quad \text{(since } D^2 - a^2 < D^2\text{)}
\]
### Step 5: Case 2 - Capacitors in Series
For the series connection, the total capacitance \( C_{eq} \) is given by:
\[
\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}
\]
Initially, the equivalent capacitance is:
\[
C_{eq, initial} = \frac{C}{2}
\]
### Step 6: Modify the Capacitor Distances in Series
Using the same modifications as before:
- For the first capacitor: \( D_1 = D - a \)
- For the second capacitor: \( D_2 = D + a \)
The new capacitances become:
\[
C_1 = \frac{\varepsilon_0 A}{D - a}
\]
\[
C_2 = \frac{\varepsilon_0 A}{D + a}
\]
### Step 7: Calculate New Equivalent Capacitance in Series
Now, the new equivalent capacitance in series is:
\[
\frac{1}{C_{eq, new}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{D - a}{\varepsilon_0 A} + \frac{D + a}{\varepsilon_0 A}
\]
Combining these gives:
\[
\frac{1}{C_{eq, new}} = \frac{(D - a) + (D + a)}{\varepsilon_0 A} = \frac{2D}{\varepsilon_0 A}
\]
Thus,
\[
C_{eq, new} = \frac{\varepsilon_0 A}{2D}
\]
This shows that:
\[
C_{eq, new} = C_{eq, initial} \quad \text{(remains constant)}
\]
### Final Results
1. In the parallel case, the total capacitance increases.
2. In the series case, the total capacitance remains constant.
