A dielectric slab is inserted between the plates of an isolatted charged capacitor. Which of the following quantities will remain the same?

Options

A. The electric field in the capacitor
B. The charge on the capacitor
C. The cpotential difference between the plates
D. The stored energy in the capacitor

Answer

To solve the problem, we need to analyze the effect of inserting a dielectric slab into an isolated charged capacitor. We will evaluate each of the given options to determine which quantity remains unchanged.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have an isolated charged capacitor, meaning it is not connected to any external voltage source (battery). The charge on the capacitor plates is fixed.

2. **Analyzing the Electric Field (E)**:
- The electric field (E) in a capacitor without a dielectric is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \(\sigma\) is the surface charge density.
- When a dielectric slab is inserted, the electric field becomes:
\[ E' = \frac{\sigma}{\epsilon_0 K} \]
where \(K\) is the dielectric constant of the material.
- Since \(K > 1\), it follows that \(E' < E\). Therefore, the electric field decreases when the dielectric is inserted. **Conclusion**: Electric field changes.

3. **Analyzing the Charge (Q)**:
- For an isolated capacitor, the charge \(Q\) on the plates remains constant because there is no external circuit to allow charge to flow. **Conclusion**: Charge remains the same.

4. **Analyzing the Potential Difference (V)**:
- The potential difference \(V\) across the capacitor is given by:
\[ V = \frac{Q}{C} \]
- When a dielectric is inserted, the capacitance increases to \(C' = K C_0\). Thus, the new potential difference becomes:
\[ V' = \frac{Q}{K C_0} \]
- Since \(K > 1\), it follows that \(V' < V\). Therefore, the potential difference decreases. **Conclusion**: Potential difference changes.

5. **Analyzing the Stored Energy (U)**:
- The energy stored in the capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
- With the new capacitance \(C' = K C_0\), the energy becomes:
\[ U' = \frac{Q^2}{2K C_0} \]
- Since \(K > 1\), it follows that \(U' < U\). Therefore, the stored energy decreases. **Conclusion**: Stored energy changes.

### Final Answer:
The quantity that remains the same when a dielectric slab is inserted between the plates of an isolated charged capacitor is **the charge on the capacitor**.

### Summary of Results:
- Electric Field: Changes
- Charge: Remains the same
- Potential Difference: Changes
- Stored Energy: Changes