A parallel plate condenser is charged by connected it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then

Options

A. potential increases
B. electric intensity increases
C. energy decreases.
D. capacity decreases

Explanation

To solve the problem step by step, we will analyze the situation of a parallel plate capacitor being charged, disconnected from a battery, and then having a glass slab introduced between the plates.

### Step 1: Understand the Initial Condition
Initially, the parallel plate capacitor is connected to a battery. When connected, it charges up to a certain voltage \( V \) and stores a charge \( Q \). The capacitance \( C \) of the capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{D} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( D \) is the separation between the plates.

**Hint:** Remember that capacitance depends on the physical characteristics of the capacitor and the medium between the plates.

### Step 2: Disconnect the Battery
Once the capacitor is fully charged, the battery is disconnected. This means that the charge \( Q \) on the capacitor remains constant because it is now an isolated system.

**Hint:** The charge remains constant when the capacitor is isolated from any external circuit.

### Step 3: Introduce the Glass Slab
When a glass slab is introduced between the plates, the capacitance of the capacitor changes. The new capacitance \( C' \) with the glass slab is given by:
\[ C' = k \cdot C = \frac{k A \epsilon_0}{D} \]
where \( k \) is the dielectric constant of the glass, which is greater than 1.

**Hint:** The introduction of a dielectric increases the capacitance of the capacitor.

### Step 4: Analyze the Effect on Voltage
Since the charge \( Q \) remains constant and the capacitance has increased, we can use the relationship between charge, capacitance, and voltage:
\[ V = \frac{Q}{C} \]
With the new capacitance \( C' \):
\[ V' = \frac{Q}{C'} \]
Since \( C' > C \), it follows that \( V' < V \). Therefore, the voltage across the capacitor decreases.

**Hint:** An increase in capacitance with constant charge results in a decrease in voltage.

### Step 5: Analyze the Electric Field
The electric field \( E \) between the plates is given by:
\[ E = \frac{V}{D} \]
Since the voltage \( V' \) decreases and the distance \( D \) remains constant, the electric field \( E' \) will also decrease:
\[ E' = \frac{V'}{D} \]

**Hint:** A decrease in voltage while keeping the distance constant leads to a decrease in electric field strength.

### Step 6: Analyze the Energy Stored
The energy \( U \) stored in the capacitor can be expressed as:
\[ U = \frac{1}{2} C V^2 \]
With the new capacitance and voltage, the energy becomes:
\[ U' = \frac{1}{2} C' V'^2 \]
Since \( C' > C \) and \( V' < V \), we can conclude that the energy stored in the capacitor decreases.

**Hint:** Energy stored in a capacitor is related to both capacitance and voltage; changes in either can affect the total energy.

### Conclusion
- The capacitance increases when the glass slab is introduced.
- The voltage across the capacitor decreases.
- The electric field strength decreases.
- The energy stored in the capacitor decreases.

### Summary of Options
1. The capacitance does not decrease; it increases.
2. The electric field decreases.
3. The energy stored decreases.

Thus, the correct conclusion is that the energy stored in the capacitor decreases.