A capacitor of `5 muF` is charged to a potential of `100 V`. Now, this charged capacitor is connected to a battery of `100 V` with the posiitive terminal of the battery connected to to the negative plate of the capacitor . For the given situation, mark the correct statement(s).

Options

A. The charge flowing through the `100 V` battery is `500 muC`,
B. The charge flowing through the `100 V` battery is `1000 muC`
C. Work done by on the battery is `0.1 J`
D. Work done on the battery is `0.1 J`.

Explanation

To solve the problem step by step, we need to analyze the situation involving the capacitor and the battery.

### Step 1: Calculate the initial charge on the capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C \) is the capacitance in farads,
- \( V \) is the voltage in volts.

Given:
- \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- \( V = 100 \, V \)

Substituting the values:
\[ Q = 5 \times 10^{-6} \, F \times 100 \, V = 5 \times 10^{-4} \, C = 500 \, \mu C \]

### Step 2: Understand the connection to the battery
When the charged capacitor is connected to a battery of the same voltage (100 V) with the positive terminal of the battery connected to the negative plate of the capacitor, the following occurs:
- The capacitor initially has a charge of \( +500 \, \mu C \) on one plate and \( -500 \, \mu C \) on the other plate.
- Connecting the battery in this manner will cause the negative plate of the capacitor to become positively charged (as it gains positive charge from the battery) and the positive plate of the capacitor to become negatively charged.

### Step 3: Calculate the final charge on the capacitor
After connecting to the battery, the capacitor will still have a capacitance of \( 5 \, \mu F \) and will be at the same voltage of \( 100 \, V \). Therefore, the final charge \( Q_f \) on the capacitor will also be:
\[ Q_f = C \times V = 5 \times 10^{-6} \, F \times 100 \, V = 500 \, \mu C \]

### Step 4: Calculate the charge that flows from the battery
The charge that flows from the battery \( Q_{flow} \) can be calculated as:
\[ Q_{flow} = Q_f - Q_i \]
where:
- \( Q_i \) is the initial charge on the capacitor, which is \( -500 \, \mu C \) (considering the negative plate),
- \( Q_f \) is the final charge on the capacitor, which is \( +500 \, \mu C \).

Thus,
\[ Q_{flow} = 500 \, \mu C - (-500 \, \mu C) = 500 \, \mu C + 500 \, \mu C = 1000 \, \mu C \]

### Step 5: Calculate the work done by the battery
The work done \( W \) by the battery is given by:
\[ W = V \times Q_{flow} \]
Substituting the values:
\[ W = 100 \, V \times 1000 \times 10^{-6} \, C = 0.1 \, J \]

### Conclusion
From the calculations, we can conclude:
1. The charge that flows from the battery is \( 1000 \, \mu C \).
2. The work done by the battery is \( 0.1 \, J \).

### Correct Statements
- The charge that flows from the battery is \( 1000 \, \mu C \) (correct).
- The work done by the battery is \( 0.1 \, J \) (correct).