A `2 mu F` condenser is charged upto 200 volt and then battery is removed. On combining this with another uncharged condenser in parallel, the potential difference is found to be 40 volt. Find the capacity of second condenser (in `mu F`)

Answer

To solve the problem step by step, we will follow the concepts of capacitors, charge, and voltage.

### Step 1: Calculate the initial charge on the first capacitor
The first capacitor has a capacitance of \( C_1 = 2 \, \mu F \) and is charged to a voltage of \( V_1 = 200 \, V \).

The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q_1 = 2 \, \mu F \times 200 \, V = 400 \, \mu C \]

### Step 2: Understand the configuration after combining with the second capacitor
After the battery is removed, the charged capacitor is connected in parallel with an uncharged capacitor (let's denote its capacitance as \( C_2 \)). The total charge remains the same, which is \( Q_1 = 400 \, \mu C \).

### Step 3: Determine the final voltage across both capacitors
After connecting the uncharged capacitor, the potential difference across both capacitors is given as \( V_f = 40 \, V \).

### Step 4: Calculate the charge on the first capacitor after combining
The charge on the first capacitor \( Q_1' \) after sharing charge can be calculated using:
\[ Q_1' = C_1 \times V_f \]
Substituting the values:
\[ Q_1' = 2 \, \mu F \times 40 \, V = 80 \, \mu C \]

### Step 5: Calculate the charge on the second capacitor
The charge on the second capacitor \( Q_2 \) can be found by subtracting the charge on the first capacitor from the total charge:
\[ Q_2 = Q_1 - Q_1' = 400 \, \mu C - 80 \, \mu C = 320 \, \mu C \]

### Step 6: Relate the charge on the second capacitor to its capacitance
The charge on the second capacitor can also be expressed as:
\[ Q_2 = C_2 \times V_f \]
Substituting the known values:
\[ 320 \, \mu C = C_2 \times 40 \, V \]

### Step 7: Solve for the capacitance of the second capacitor
Rearranging the equation to find \( C_2 \):
\[ C_2 = \frac{320 \, \mu C}{40 \, V} = 8 \, \mu F \]

### Final Answer
The capacitance of the second capacitor is \( C_2 = 8 \, \mu F \).

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