Half of the space between the plates of a parallel plate capacitor is filled with dielectric material of constant K. Then which of the plots are possible?

Options

A.
B.
C.
D.

Explanation

To solve the problem of a parallel plate capacitor with half of the space filled with a dielectric material of constant \( K \), we will analyze the electric field and potential difference across the capacitor.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- A parallel plate capacitor consists of two plates separated by a distance \( d \).
- In this case, half of the space (i.e., \( d/2 \)) is filled with a dielectric material of constant \( K \), and the other half is air (or vacuum).

2. **Electric Field in Different Regions**:
- Let \( E \) be the electric field in the air gap (the region without the dielectric).
- In the dielectric region, the electric field \( E' \) can be expressed as:
\[ E' = \frac{E}{K} \]
- This is because the presence of the dielectric reduces the electric field strength by a factor of \( K \).

3. **Potential Difference Calculation**:
- The potential difference \( V \) across the capacitor can be calculated by integrating the electric field over the distance.
- For the air gap (first half):
\[ V_{\text{air}} = E \cdot \frac{d}{2} \]
- For the dielectric (second half):
\[ V_{\text{dielectric}} = E' \cdot \frac{d}{2} = \frac{E}{K} \cdot \frac{d}{2} \]
- Therefore, the total potential difference \( V \) across the capacitor is:
\[ V = V_{\text{air}} + V_{\text{dielectric}} = E \cdot \frac{d}{2} + \frac{E}{K} \cdot \frac{d}{2} \]
- Simplifying this gives:
\[ V = \frac{dE}{2} \left(1 + \frac{1}{K}\right) \]

4. **Graphical Representation**:
- For the electric field \( E \) versus position \( x \):
- In the air gap, the electric field \( E \) is constant.
- In the dielectric, the electric field \( E' \) is constant but lower than \( E \).
- Therefore, the graph of \( E \) versus \( x \) will show a step down at the boundary of the dielectric.
- For the potential difference \( V \) versus position \( x \):
- The slope of \( V \) in the dielectric region will be less steep than in the air gap due to the reduced electric field.
- Hence, the graph of \( V \) versus \( x \) will show a steeper slope in the air gap and a gentler slope in the dielectric.

5. **Conclusion**:
- The possible plots for the electric field and potential difference are:
- Electric field \( E \) versus position \( x \): a step function with a drop at the dielectric boundary.
- Potential difference \( V \) versus position \( x \): a linear function with a steeper slope in the air gap and a gentler slope in the dielectric.

### Final Answer:
The possible plots are:
- Electric field \( E \) versus position \( x \): Option D (step function).
- Potential difference \( V \) versus position \( x \): Option A (linear function with varying slope).