A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

Options

A. increase
B. decrease
C. remain unchanged
D. become zero

Answer

To solve the problem of how the force between the plates of an isolated capacitor changes when a dielectric slab is inserted, we can follow these steps:

### Step 1: Understand the Initial Conditions
An isolated capacitor has a fixed charge (Q) on its plates. The force between the plates of a capacitor can be expressed using the formula:
\[ F = \frac{Q^2}{2A \epsilon_0} \]
where:
- \( F \) is the force between the plates,
- \( Q \) is the charge on the plates,
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Insert the Dielectric Slab
When a dielectric slab is inserted between the plates of the capacitor, it affects the electric field between the plates. However, since the capacitor is isolated, the charge (Q) on the plates remains constant.

### Step 3: Analyze the Effect of the Dielectric
The presence of the dielectric slab increases the capacitance of the capacitor. The new capacitance \( C' \) can be expressed as:
\[ C' = K \cdot C \]
where \( K \) is the dielectric constant of the material and \( C \) is the original capacitance without the dielectric.

### Step 4: Determine the New Force
Even though the capacitance increases, the charge remains constant. The force between the plates can still be expressed using the original formula:
\[ F = \frac{Q^2}{2A \epsilon_0} \]
Since \( Q \) and \( A \) are constant and \( \epsilon_0 \) does not change, the force remains unchanged.

### Conclusion
Thus, the force between the plates of an isolated capacitor with a dielectric slab inserted remains the same.

### Final Answer
The force between the plates will remain unchanged.

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