A parallel-plate capacitor is filled with a dielectric up to one-half of the distance between the plates.The manner in which the potential between the plates varies is illustrated in the figure. Which half `(1 or 2)` of the space between the plates is filled with the dielectric and what will be the distribution of the potential after the dielectric is taken out of the capacitor provided that (a) the charges on the plates are conserved or (b) the potential difference across the capacitor is conserved?

Answer

To solve the problem, we need to analyze the behavior of a parallel-plate capacitor filled with a dielectric and how the potential distribution changes when the dielectric is removed under two different conditions: (a) when the charge on the plates is conserved and (b) when the potential difference across the capacitor is conserved.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a parallel-plate capacitor with plates separated by a distance \(d\).
- A dielectric material fills half of the space between the plates (let's assume it fills the lower half).
- The dielectric has a dielectric constant \(K\), which reduces the electric field and the potential difference across that region.

2. **Potential Distribution with Dielectric**:
- The potential difference \(V\) across the capacitor can be expressed as:
\[ V = V_1 + V_2 \]
where \(V_1\) is the potential across the dielectric and \(V_2\) is the potential across the air gap.
- The potential across the dielectric can be expressed as:
\[ V_1 = \frac{Q}{K \cdot A} \cdot \frac{d/2}{\epsilon_0} \]
and across the air gap:
\[ V_2 = \frac{Q}{A} \cdot \frac{d/2}{\epsilon_0} \]
- The total potential \(V\) will be lower than it would be without the dielectric due to the presence of \(K\).

3. **Case (a): Charge on the Plates is Conserved**:
- When the dielectric is removed, the charge \(Q\) on the plates remains constant.
- The potential distribution will change because the dielectric no longer reduces the electric field.
- The new potential distribution will be linear across the entire distance \(d\) and can be represented as:
\[ V' = \frac{Q}{A \cdot \epsilon_0} \cdot d \]
- The graph of potential distribution will be a straight line from 0 to \(V'\), indicating a uniform electric field throughout.

4. **Case (b): Potential Difference Across the Capacitor is Conserved**:
- When the dielectric is removed and the potential difference \(V\) is conserved, the total potential difference remains the same.
- However, the distribution will still be linear, but it will be at a lower level than the original potential distribution with the dielectric.
- The new potential distribution will look similar to the case where charge is conserved, but it will be slightly lower than the previous graph, indicating that the potential difference is maintained but the electric field is uniform.

5. **Conclusion**:
- In both cases, after the dielectric is removed, the potential distribution becomes linear.
- The first half (where the dielectric was) will behave like the second half (where there was air), leading to a uniform potential across the capacitor.