A capacitor when filled with a dielectric `K = 3` has charge `Q_(0)`, voltage `V_(0)` and filed `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)`

Answer

To solve the problem, we need to analyze the changes in the capacitor's properties when the dielectric material is changed from \( K = 3 \) to \( K = 9 \). We will denote the initial values with subscript 0 and the new values with subscript 1.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Charge, Voltage, and Capacitance:**
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

2. **Capacitance with Dielectric:**
The capacitance of a capacitor filled with a dielectric material is given by:
\[ C = K \cdot C_0 \]
where \( K \) is the dielectric constant and \( C_0 \) is the capacitance without the dielectric.

3. **Initial Conditions:**
For the initial dielectric with \( K = 3 \):
\[ C_0 = C_0 \quad (initial \, capacitance) \]
\[ C_1 = 3 \cdot C_0 \quad (new \, capacitance \, with \, K = 3) \]
The initial charge and voltage are:
\[ Q_0 = C_0 \cdot V_0 \]

4. **New Conditions with Dielectric \( K = 9 \):**
When the dielectric is replaced with one having \( K = 9 \):
\[ C_2 = 9 \cdot C_0 \quad (new \, capacitance \, with \, K = 9) \]

5. **Charge Conservation:**
Since there is no battery connected, the charge remains constant:
\[ Q_1 = Q_0 \]

6. **Finding New Voltage \( V_1 \):**
Using the charge conservation:
\[ Q_1 = C_2 \cdot V_1 \]
Substituting \( Q_0 \) for \( Q_1 \):
\[ Q_0 = (9 \cdot C_0) \cdot V_1 \]
Now substituting \( Q_0 = C_0 \cdot V_0 \):
\[ C_0 \cdot V_0 = (9 \cdot C_0) \cdot V_1 \]
Dividing both sides by \( C_0 \) (assuming \( C_0 \neq 0 \)):
\[ V_0 = 9 \cdot V_1 \]
Thus,
\[ V_1 = \frac{V_0}{9} \]

7. **Finding New Electric Field \( E_1 \):**
The electric field \( E \) is related to voltage and distance \( d \) by:
\[ E = \frac{V}{d} \]
Since the distance \( d \) remains constant, the electric field will change proportionally to the voltage:
\[ E_1 = \frac{V_1}{d} = \frac{V_0/9}{d} = \frac{E_0}{9} \]

### Final Results:
- New Charge \( Q_1 = Q_0 \)
- New Voltage \( V_1 = \frac{V_0}{9} \)
- New Electric Field \( E_1 = \frac{E_0}{9} \)

### Summary:
The new values of charge, voltage, and electric field when the dielectric is replaced with \( K = 9 \) are:
- Charge: \( Q_1 = Q_0 \)
- Voltage: \( V_1 = \frac{V_0}{9} \)
- Electric Field: \( E_1 = \frac{E_0}{9} \)