Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential differences across the capacitors now becomes...........

Options

A. `(2V)/(K+2)`
B. `V/(K+2)`
C. `(3V)/(K+3)`
D. `(3V)/(K+2)`

Explanation

To solve the problem step by step, we will analyze the situation involving the two capacitors, their initial conditions, and the effect of inserting a dielectric into one of them.

### Step 1: Understand the initial conditions
We have two capacitors connected in parallel:
- Capacitor 1 has capacitance \( C \)
- Capacitor 2 has capacitance \( 2C \)

Both capacitors are charged to a potential difference \( V \).

**Hint:** Remember that when capacitors are connected in parallel, the total charge is the sum of the charges on each capacitor.

### Step 2: Calculate the total initial charge
The charge on each capacitor can be calculated using the formula \( Q = CV \):
- Charge on Capacitor 1: \( Q_1 = C \cdot V \)
- Charge on Capacitor 2: \( Q_2 = 2C \cdot V \)

The total charge \( Q_T \) in the system is:
\[ Q_T = Q_1 + Q_2 = CV + 2CV = 3CV \]

**Hint:** The total charge remains constant when the battery is disconnected.

### Step 3: Insert the dielectric
When the dielectric with a dielectric constant \( K \) is inserted into Capacitor 1 (capacitance \( C \)), its capacitance changes to:
\[ C' = K \cdot C \]

**Hint:** The dielectric increases the capacitance of the capacitor, which will affect the potential difference across it.

### Step 4: Define the new potential difference
Let the new potential difference across the capacitors be \( V' \). The charges on the capacitors after the dielectric is inserted will be:
- Charge on Capacitor 1 (with dielectric): \( Q_1' = K \cdot C \cdot V' \)
- Charge on Capacitor 2 (unchanged): \( Q_2' = 2C \cdot V' \)

### Step 5: Apply charge conservation
Since the total charge remains conserved, we can write:
\[ Q_T = Q_T' \implies 3CV = Q_1' + Q_2' = KCV' + 2CV' \]

**Hint:** Set up the equation based on the conservation of charge.

### Step 6: Rearrange the equation
Now we can rearrange the equation:
\[ 3CV = (K + 2)CV' \]
Dividing through by \( C \) (assuming \( C \neq 0 \)):
\[ 3V = (K + 2)V' \]

### Step 7: Solve for the new potential difference
Now, we can solve for \( V' \):
\[ V' = \frac{3V}{K + 2} \]

**Hint:** This expression gives the new potential difference across the capacitors after the dielectric is inserted.

### Final Answer
The potential difference across the capacitors after inserting the dielectric is:
\[ V' = \frac{3V}{K + 2} \]