On placing dielectric slab between the plates of an isolated charged condenser its

Options

A. `{:("Capcitance Potential Difference Energy stored Electric Field"),("decreases decreases increases increases"):}`
B. `{:("Capcitance Potential Difference Energy stored Electric Field"),("increases increases increases decreases"):}`
C. `{:("Capcitance Potential Difference Energy stored Electric Field"),("increases decreases decreases decreases"):}`
D. `{:("Capcitance Potential Difference Energy stored Electric Field"),("decreases decreases increases remains"):}`

Answer

To solve the problem of what happens when a dielectric slab is placed between the plates of an isolated charged capacitor, we will analyze the effects on capacitance, potential difference, energy, and electric field step by step.

### Step 1: Understand the Initial Conditions
- An isolated charged capacitor means that the charge (Q) on the capacitor remains constant when the dielectric slab is introduced.
- The initial capacitance (C₀) of the capacitor without the dielectric is given by the formula:
\[ C₀ = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

**Hint:** Remember that in an isolated capacitor, the charge remains constant.

### Step 2: Effect of Introducing the Dielectric
- When a dielectric slab with dielectric constant \( K \) and thickness \( t \) is placed between the plates, the new capacitance (C) can be calculated using the formula:
\[ C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} \]
- Since \( K > 1 \), the term \( \frac{t}{K} \) is less than \( t \), which means the effective distance between the plates is reduced. Thus, \( d - t + \frac{t}{K} < d \).

**Hint:** The introduction of a dielectric increases the capacitance because it effectively reduces the distance between the plates.

### Step 3: Analyze the Change in Capacitance
- Since the capacitance increases (C > C₀), we can conclude that:
\[ C > C₀ \]

**Hint:** An increase in capacitance occurs because the dielectric reduces the effective distance between the plates.

### Step 4: Relationship Between Charge, Capacitance, and Voltage
- The relationship between charge, capacitance, and voltage is given by:
\[ Q = C \cdot V \]
- Since the charge (Q) is constant and the capacitance (C) has increased, the potential difference (V) must decrease to maintain the equality:
\[ V = \frac{Q}{C} \]
- Therefore, if C increases, V must decrease.

**Hint:** If capacitance increases while charge remains constant, the voltage must decrease.

### Step 5: Analyze the Change in Energy
- The energy (U) stored in the capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
- Since Q is constant and C has increased, the energy stored in the capacitor must decrease:
\[ U < U₀ \]

**Hint:** Energy decreases when capacitance increases while charge remains constant.

### Step 6: Analyze the Change in Electric Field
- The electric field (E) between the plates of a capacitor is related to the potential difference and the distance between the plates:
\[ E = \frac{V}{d} \]
- Since the potential difference (V) has decreased and the distance (d) remains the same, the electric field (E) must also decrease.

**Hint:** A decrease in potential difference leads to a decrease in electric field strength.

### Conclusion
In summary, when a dielectric slab is placed between the plates of an isolated charged capacitor:
1. The capacitance increases.
2. The potential difference decreases.
3. The energy stored in the capacitor decreases.
4. The electric field strength decreases.

The correct option that summarizes these changes is the third option, which states that the capacitance increases, the potential difference decreases, and the energy decreases.