`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` volts
B. `120` volts
C. `300` volts
D. `600` volts

Explanation

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

### Step 1: Calculate the initial charge on the capacitor
The initial charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \cdot V \]
where:
- \( C = 0.2 \, \text{F} \) (capacitance of the first capacitor)
- \( V = 600 \, \text{V} \) (voltage across the first capacitor)

Substituting the values:
\[ Q = 0.2 \, \text{F} \cdot 600 \, \text{V} = 120 \, \text{C} \]

### Step 2: Understand the connection of the second capacitor
When the charged capacitor is connected to the second capacitor (1 F), they are in parallel. In parallel, the voltage across both capacitors will be the same after they are connected.

### Step 3: Set up the equations for charge conservation
Let \( Q_1 \) be the charge on the 0.2 F capacitor and \( Q_2 \) be the charge on the 1 F capacitor after they are connected. The total charge before connecting the capacitors must equal the total charge after they are connected:
\[ Q_1 + Q_2 = 120 \, \text{C} \]

### Step 4: Relate the charges to the voltages
The voltage across each capacitor after connection will be the same, denoted as \( V_f \). Therefore, we can express the charges as:
\[ Q_1 = C_1 \cdot V_f = 0.2 \cdot V_f \]
\[ Q_2 = C_2 \cdot V_f = 1 \cdot V_f \]

### Step 5: Substitute into the charge conservation equation
Substituting these expressions into the charge conservation equation:
\[ 0.2 V_f + 1 V_f = 120 \]
\[ 1.2 V_f = 120 \]

### Step 6: Solve for the final voltage \( V_f \)
Now, solve for \( V_f \):
\[ V_f = \frac{120}{1.2} = 100 \, \text{V} \]

### Conclusion
The potential across the capacitors after connecting them will decrease to:
\[ \boxed{100 \, \text{V}} \]

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