A dielectric is inserted between the plates of an isolated fully-charged capacitor. The dielectric completely fills the space between the plates. The magnitude of electrostatic force on either metal plate decreases, as it was before the insertion of dielectric medium.
Due to insertion of dielectric slab in an isolated parallel plate capacitor (the dielectric completely fills the space between the plates), the electrostatic potential energy of the capacitor decreases.

Options

A. Both A and R false
B. Both A and R true and R is not correct reason for A
C. A is true and R is false
D. Both A and R are true and R is correct reason of A.

Answer

To solve the problem, we need to analyze the effects of inserting a dielectric between the plates of an isolated, fully charged capacitor. We will consider the changes in electrostatic force and potential energy.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have an isolated fully charged capacitor with charge \( Q \) and capacitance \( C \).
- The initial electric field \( E \) between the plates is given by \( E = \frac{V}{d} \), where \( V \) is the voltage across the capacitor and \( d \) is the distance between the plates.

2. **Force on the Plates**:
- The electrostatic force \( F \) on one plate of the capacitor can be expressed as:
\[ F = E \cdot Q \]
- Initially, the force is \( F_0 = E \cdot Q \).

3. **Inserting the Dielectric**:
- When a dielectric material with dielectric constant \( K \) is inserted, the electric field \( E' \) between the plates becomes:
\[ E' = \frac{E}{K} \]
- The new force \( F' \) on the plates after inserting the dielectric is:
\[ F' = E' \cdot Q = \frac{E}{K} \cdot Q \]
- Since \( K > 1 \), it follows that \( F' < F_0 \). Thus, the magnitude of the electrostatic force on either plate decreases.

4. **Potential Energy of the Capacitor**:
- The potential energy \( U \) of the capacitor before inserting the dielectric is given by:
\[ U_0 = \frac{Q^2}{2C} \]
- After inserting the dielectric, the capacitance increases to \( C' = K \cdot C \). The new potential energy \( U' \) is:
\[ U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(K \cdot C)} = \frac{U_0}{K} \]
- Since \( K > 1 \), it follows that \( U' < U_0 \). Thus, the electrostatic potential energy of the capacitor decreases.

5. **Conclusion**:
- Both statements in the question are correct: the electrostatic force on either plate decreases, and the electrostatic potential energy of the capacitor also decreases.