A capacitor when filled with a dielectric `K = 3` has charge `Q_(0)`, voltage `V_(0)` and field `E_(0)`. If the dielectric is replaced with another one having `K = 9`, the new values of charge, voltage and field will be respectively

Options

A. `3Q_(0), 3V_(0), 3E_(0)`
B. `Q_(0), 3V_(0), 3E_(0)`
C. `Q_(0), (V_(0))/(3), 3E_(0)`
D. `Q_(0), (V_(0))/(3), (E_(0))/(3)`

Explanation

To solve the problem step by step, we will analyze the effects of replacing the dielectric in the capacitor and derive the new values of charge, voltage, and electric field.

### Step 1: Understand the initial conditions
- The initial dielectric constant \( K_1 = 3 \).
- The initial charge on the capacitor \( Q_0 \).
- The initial voltage across the capacitor \( V_0 \).
- The initial electric field \( E_0 \).

### Step 2: Determine the capacitance with the initial dielectric
The capacitance \( C_1 \) of a capacitor filled with a dielectric is given by:
\[ C_1 = K_1 \cdot C_0 \]
where \( C_0 \) is the capacitance without the dielectric. Thus,
\[ C_1 = 3 \cdot C_0 \]

### Step 3: Analyze the situation when the dielectric is replaced
When the dielectric is replaced with another one having \( K_2 = 9 \), the new capacitance \( C_2 \) becomes:
\[ C_2 = K_2 \cdot C_0 = 9 \cdot C_0 \]

### Step 4: Conservation of charge
Since the battery is disconnected, the charge on the capacitor remains constant. Therefore, the charge after replacing the dielectric is still:
\[ Q_2 = Q_0 \]

### Step 5: Relate charge, capacitance, and voltage
The relationship between charge, capacitance, and voltage is given by:
\[ Q = C \cdot V \]
For the initial conditions:
\[ Q_0 = C_1 \cdot V_0 \]
For the new conditions:
\[ Q_2 = C_2 \cdot V_2 \]
Since \( Q_2 = Q_0 \), we have:
\[ Q_0 = C_2 \cdot V_2 \]

### Step 6: Solve for the new voltage \( V_2 \)
Substituting \( C_2 \) into the equation:
\[ Q_0 = (9 \cdot C_0) \cdot V_2 \]
From the initial condition, we know \( Q_0 = (3 \cdot C_0) \cdot V_0 \). Thus, we can equate:
\[ (3 \cdot C_0) \cdot V_0 = (9 \cdot C_0) \cdot V_2 \]
Cancelling \( C_0 \) (assuming it is not zero):
\[ 3 \cdot V_0 = 9 \cdot V_2 \]
Solving for \( V_2 \):
\[ V_2 = \frac{3}{9} \cdot V_0 = \frac{1}{3} V_0 \]

### Step 7: Calculate the new electric field \( E_2 \)
The electric field \( E \) in a capacitor is related to the voltage and the distance \( d \) between the plates:
\[ E = \frac{V}{d} \]
Thus, the new electric field \( E_2 \) can be calculated as:
\[ E_2 = \frac{V_2}{d} = \frac{\frac{1}{3} V_0}{d} = \frac{1}{3} E_0 \]

### Summary of Results
- The new charge \( Q_2 = Q_0 \)
- The new voltage \( V_2 = \frac{1}{3} V_0 \)
- The new electric field \( E_2 = \frac{1}{3} E_0 \)

### Final Answer
The new values of charge, voltage, and electric field are:
- Charge: \( Q_0 \)
- Voltage: \( \frac{1}{3} V_0 \)
- Electric Field: \( \frac{1}{3} E_0 \)