Consider these three statements for a capacitor
STATEMENT-1 `:` Capacitance of a capacitor must increase when a dielectric is inserted in between the capacitor plates.
STATEMENT-2 `:` Potential differenece across a charged capacitor must increase if a dielectric is inserted in between the capacitor plates.
STATEMENT-3 `:` Electrostatic energy stored by the capacitor may increase if a dielectric inserted in between the capacitor plates

Options

A. T F F
B. F T F
C. T F T
D. F T T

Answer

To analyze the three statements regarding a capacitor, we will evaluate each statement step by step.

### Step 1: Evaluate Statement 1
**Statement 1:** Capacitance of a capacitor must increase when a dielectric is inserted in between the capacitor plates.

**Solution:**
- The capacitance \( C \) of a capacitor without a dielectric is given by the formula:
\[ C_0 = \frac{\epsilon_0 A}{d} \]
where \( A \) is the area of the plates, \( d \) is the distance between them, and \( \epsilon_0 \) is the permittivity of free space.
- When a dielectric material with dielectric constant \( k \) is inserted, the new capacitance \( C \) becomes:
\[ C = k \cdot C_0 = k \cdot \frac{\epsilon_0 A}{d} \]
- Since \( k > 1 \), it follows that \( C > C_0 \). Therefore, the capacitance increases when a dielectric is inserted.

**Conclusion:** Statement 1 is **True**.

### Step 2: Evaluate Statement 2
**Statement 2:** Potential difference across a charged capacitor must increase if a dielectric is inserted in between the capacitor plates.

**Solution:**
- Consider a charged capacitor with charge \( Q \) and initial capacitance \( C_0 \). The initial potential difference \( V_0 \) is given by:
\[ V_0 = \frac{Q}{C_0} \]
- When a dielectric is inserted and if the capacitor is isolated (battery removed), the charge \( Q \) remains constant, but the capacitance changes to \( C = k \cdot C_0 \).
- The new potential difference \( V' \) becomes:
\[ V' = \frac{Q}{C} = \frac{Q}{k \cdot C_0} = \frac{V_0}{k} \]
- Since \( k > 1 \), it follows that \( V' < V_0 \). Therefore, the potential difference decreases when a dielectric is inserted.

**Conclusion:** Statement 2 is **False**.

### Step 3: Evaluate Statement 3
**Statement 3:** Electrostatic energy stored by the capacitor may increase if a dielectric is inserted in between the capacitor plates.

**Solution:**
- The electrostatic energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
- If the battery remains connected while inserting the dielectric, the capacitance changes to \( C = k \cdot C_0 \) and the potential \( V \) remains constant.
- The new energy \( U' \) becomes:
\[ U' = \frac{1}{2} (k \cdot C_0) V^2 = k \cdot \left(\frac{1}{2} C_0 V^2\right) = k \cdot U \]
- Since \( k > 1 \), it follows that \( U' > U \). Therefore, the energy stored increases when a dielectric is inserted, provided the battery remains connected.

**Conclusion:** Statement 3 is **True**.

### Final Summary
- Statement 1: True
- Statement 2: False
- Statement 3: True

Thus, the correct answer is that Statement 1 and Statement 3 are true, while Statement 2 is false.

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