The plates of a parallel plate capacitor with no dielectirc are connected to a volatage source. Now a dielectric of dielectric constant `K` is inserted to fill the whose space between the plates with voltage source remainign connected to the capacitor-

Options

A. The energy stored in the capacitor will become will decrease k times
B. The electric field of inside the capacitor will decrease k times
C. The force of attration between the plates will become `K^(2)` times
D. The charge on the capacitor will become will k times

Answer

To solve the problem step by step, we will analyze the effects of inserting a dielectric into a parallel plate capacitor that is connected to a voltage source.

### Step 1: Understanding the Initial Condition
- Initially, we have a parallel plate capacitor with no dielectric connected to a voltage source of \( V \) volts.
- The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( A \) is the area of the plates, \( d \) is the separation between the plates, and \( \varepsilon_0 \) is the permittivity of free space.

### Step 2: Charge on the Capacitor
- The charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \cdot V \]
- Therefore, the initial charge on the capacitor is:
\[ Q_1 = C \cdot V \]

### Step 3: Inserting the Dielectric
- When a dielectric with dielectric constant \( K \) is inserted, the capacitance increases to:
\[ C' = K \cdot C \]
- Since the capacitor remains connected to the voltage source, the voltage \( V \) across the capacitor remains constant.

### Step 4: New Charge on the Capacitor
- The new charge \( Q' \) on the capacitor after inserting the dielectric is:
\[ Q' = C' \cdot V = K \cdot C \cdot V = K \cdot Q_1 \]
- This shows that the charge on the capacitor increases by a factor of \( K \).

### Step 5: Energy Stored in the Capacitor
- The energy stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
- The initial energy \( E_1 \) is:
\[ E_1 = \frac{1}{2} C V^2 \]
- The new energy \( E_2 \) after inserting the dielectric becomes:
\[ E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} (K \cdot C) V^2 = K \cdot \left(\frac{1}{2} C V^2\right) = K \cdot E_1 \]
- This indicates that the energy stored in the capacitor increases by a factor of \( K \).

### Step 6: Electric Field Inside the Capacitor
- The electric field \( E \) inside a capacitor is given by:
\[ E = \frac{V}{d} \]
- Since both \( V \) and \( d \) remain constant when the dielectric is inserted, the electric field \( E \) does not change. Therefore, it remains constant.

### Step 7: Force of Attraction Between the Plates
- The force \( F \) of attraction between the plates of a capacitor can be expressed as:
\[ F = \frac{Q^2}{2A \varepsilon_0} \]
- Substituting \( Q' = K \cdot Q_1 \) into the force equation gives:
\[ F' = \frac{(K \cdot Q_1)^2}{2A \varepsilon_0} = K^2 \cdot \frac{Q_1^2}{2A \varepsilon_0} = K^2 \cdot F_1 \]
- This shows that the force of attraction increases by a factor of \( K^2 \).

### Conclusion
- The effects of inserting a dielectric into the capacitor while connected to a voltage source are:
1. The energy stored in the capacitor increases by a factor of \( K \).
2. The electric field inside the capacitor remains constant.
3. The force of attraction between the plates increases by a factor of \( K^2 \).
4. The charge on the capacitor increases by a factor of \( K \).

### Summary of Options
- Option 1: The energy stored in the capacitor will decrease \( K \) times. **(Incorrect)**
- Option 2: The electric field inside the capacitor will decrease \( K \) times. **(Incorrect)**
- Option 3: The force of attraction between the plates becomes \( K^2 \) times. **(Correct)**
- Option 4: The charge on the capacitor will become \( K \) times. **(Correct)**