Consider a parallel plate capacitor originally with a charge `q_0` capacitance `C_0` and potential difference `Delta V_0` There is an electrostatic force of magnitude `F_0` between the plates and the capacitor has a stored energy `U_0`. The terminals of the capacitor are connected to another capacitor of same capacitance and charge.
Later the dielectric slab is removed. WHile the slab is being removed.

Options

A. Charge on second capacitor increases
B. Charge on second capacitor decreases
C. Charge on second capacitor is constant
D. Can't be said

Answer

To solve the problem step by step, we will analyze the situation involving the parallel plate capacitor, the connection to another capacitor, and the removal of the dielectric slab.

### Step 1: Understand the Initial Conditions
We have a parallel plate capacitor with:
- Charge: \( q_0 \)
- Capacitance: \( C_0 \)
- Potential difference: \( \Delta V_0 \)
- Electrostatic force: \( F_0 \)
- Stored energy: \( U_0 \)

### Step 2: Analyze the Connection to Another Capacitor
The terminals of the capacitor are connected to another capacitor of the same capacitance \( C_0 \) and charge \( q_0 \). When two capacitors are connected in parallel, the total capacitance \( C_{\text{total}} \) becomes:
\[ C_{\text{total}} = C_0 + C_0 = 2C_0 \]
The total charge \( Q_{\text{total}} \) on the system remains the same as the charge on the first capacitor:
\[ Q_{\text{total}} = q_0 + q_0 = 2q_0 \]

### Step 3: Effect of Removing the Dielectric Slab
When the dielectric slab is removed from the first capacitor, the capacitance \( C \) of the capacitor changes. The capacitance without the dielectric is given by:
\[ C = \frac{C_0}{\kappa} \]
where \( \kappa \) is the dielectric constant.

### Step 4: Analyze the Changes in Charge and Voltage
When the dielectric is removed, the charge \( q \) on the capacitor will change due to the change in capacitance. The relationship between charge, capacitance, and voltage is given by:
\[ q = C \Delta V \]
Since the charge on the capacitor remains constant while the dielectric is being removed, the potential difference will change.

### Step 5: Determine the New Values
1. **New Capacitance**: After removing the dielectric, the capacitance becomes \( C = \frac{C_0}{\kappa} \).
2. **New Voltage**: The new voltage across the capacitor can be calculated using the relationship:
\[ \Delta V = \frac{q_0}{C} \]
Substituting for \( C \):
\[ \Delta V = \frac{q_0 \kappa}{C_0} \]

### Step 6: Conclusion
As the dielectric slab is removed:
- The capacitance decreases.
- The voltage across the capacitor increases.
- The charge on the capacitor remains constant during the removal process.