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.
A dielectric slab with `k_e gt 1` is inserted between the plates of the first capacitor . Which quantity decreases?

Options

A. q
B. C
C. `Delta V`
D. F

Answer

To solve the problem, we will analyze the effect of inserting a dielectric slab into a parallel plate capacitor that is connected to another capacitor of the same capacitance and charge.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have a parallel plate capacitor with initial charge \( q_0 \), capacitance \( C_0 \), and potential difference \( \Delta V_0 \).
- The electrostatic force between the plates is \( F_0 \), and the stored energy is \( U_0 \).

2. **Connecting Another Capacitor**:
- The terminals of the first 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 becomes \( C_{\text{total}} = C_0 + C_0 = 2C_0 \).
- The total charge \( Q \) remains the same, but the voltage across the combined system will change.

3. **Inserting the Dielectric Slab**:
- When a dielectric slab with dielectric constant \( k_e > 1 \) is inserted into the first capacitor, the capacitance of the first capacitor increases.
- The new capacitance \( C' \) of the first capacitor becomes \( C' = k_e C_0 \).

4. **Effect on Voltage**:
- The relationship between charge, capacitance, and voltage is given by \( Q = C \Delta V \).
- Since the charge \( q_0 \) remains constant, we can write:
\[ q_0 = C' \Delta V' \quad \text{(after inserting the dielectric)} \]
- Substituting \( C' \):
\[ q_0 = k_e C_0 \Delta V' \]
- Rearranging gives:
\[ \Delta V' = \frac{q_0}{k_e C_0} \]
- Since \( k_e > 1 \), it follows that \( \Delta V' < \Delta V_0 \). Thus, the potential difference decreases.

5. **Conclusion**:
- The quantity that decreases when the dielectric slab is inserted is the potential difference \( \Delta V \).

### Final Answer:
The potential difference \( \Delta V \) decreases.