Given information:
- Two capacitors are connected in series.
- One capacitor has a capacitance of 10 microfarads (µF) and the other has a capacitance of 20 µF.
- The capacitors are connected across a 200 volt supply line.
- The charged capacitors are then disconnected from the line and reconnected with positive plates together and negative plates together.
- An external voltage is applied.

Explanation:

Step 1: Capacitors in series
When capacitors are connected in series, their effective capacitance (C) can be calculated using the formula:

1/C = 1/C1 + 1/C2 + 1/C3 + ...

In this case, we have two capacitors in series, so the effective capacitance is given by:

1/C = 1/10 µF + 1/20 µF

Simplifying this equation:

1/C = (2 + 1)/20 µF

1/C = 3/20 µF

C = 20/3 µF

Therefore, the effective capacitance of the series combination is 20/3 µF.

Step 2: Charging the capacitors
The series combination of capacitors is connected across a 200 volt supply line. Since the capacitors are in series, the potential difference (V) across each capacitor will be the same.

Using the formula for capacitance:

C = Q/V

Where:
C is the capacitance,
Q is the charge stored on the capacitor, and
V is the potential difference across the capacitor.

We can rearrange this formula to solve for V:

V = Q/C

Since the capacitors are connected in series, the charge stored on each capacitor will be the same. Let's assume the charge stored on each capacitor is Q.

For the 10 µF capacitor:

V1 = Q/(10 µF) (Equation 1)

For the 20 µF capacitor:

V2 = Q/(20 µF) (Equation 2)

Since V1 = V2 (the potential difference across each capacitor is the same), we can equate Equations 1 and 2:

Q/(10 µF) = Q/(20 µF)

Simplifying this equation:

1/10 µF = 1/20 µF

This equation is not possible. Therefore, our assumption that the charge stored on each capacitor is the same is incorrect.

Step 3: Reconnecting the capacitors
When the charged capacitors are disconnected from the line and reconnected with positive plates together and negative plates together, the charges on the capacitors redistribute. The charge Q on each capacitor remains the same, but the potential difference across each capacitor changes.

Step 4: Applying an external voltage
An external voltage is applied to the capacitors. Since the potential difference across each capacitor is the same, the external voltage will be divided between the capacitors based on their capacitances.

The potential difference across the 10 µF capacitor can be calculated using the formula:

V1 = Q/(10 µF)

The potential difference across the 20 µF capacitor can be calculated using the formula:

V2 = Q/(20 µF)

Since the charge Q remains the same, the potential difference across each capacitor will be inversely proportional to their capacit