`0.2 F` capacitor is charged to `600 V` by a battery. On removing the battery it is connected with another parallel plate condenser of `1 F`. The potential decreases to

Options

A. 100 V
B. 120 V
C. 300 V
D. 600 V

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial charge on the 0.2 F capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

For the 0.2 F capacitor charged to 600 V:
\[ Q = 0.2 \, \text{F} \times 600 \, \text{V} = 120 \, \text{C} \]

### Step 2: Understand the connection with the 1 F capacitor
When the charged 0.2 F capacitor is connected to the 1 F capacitor, the total charge will redistribute between the two capacitors. The total charge remains conserved, which means:
\[ Q_{\text{total}} = Q_1 + Q_2 \]
where \( Q_1 \) is the charge on the 0.2 F capacitor and \( Q_2 \) is the charge on the 1 F capacitor.

### Step 3: Set up the equations for charge
Let \( V_{AB} \) be the final voltage across both capacitors after they are connected. The charge on each capacitor can be expressed as:
\[ Q_1 = C_1 \times V_{AB} = 0.2 \, \text{F} \times V_{AB} \]
\[ Q_2 = C_2 \times V_{AB} = 1 \, \text{F} \times V_{AB} \]

### Step 4: Write the conservation of charge equation
From the conservation of charge:
\[ Q_{\text{total}} = Q_1 + Q_2 \]
Substituting the values:
\[ 120 \, \text{C} = (0.2 \, \text{F} \times V_{AB}) + (1 \, \text{F} \times V_{AB}) \]
This simplifies to:
\[ 120 = 0.2 V_{AB} + 1 V_{AB} \]
\[ 120 = 1.2 V_{AB} \]

### Step 5: Solve for \( V_{AB} \)
Now we can solve for \( V_{AB} \):
\[ V_{AB} = \frac{120}{1.2} = 100 \, \text{V} \]

### Final Answer
The potential decreases to \( 100 \, \text{V} \).

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