A parallel plate capacitor of plate area A and plates separation distance d is charged by applying a potential `V_(0)` between the plates. The dielectric constant of the medium between the plates is K. What is the uniform electric field E between the plates of the capacitor ?

Options

A. `E^(2)Ad//epsilon_(0)`
B. `(1)/(2)epsilon_(0)E^(2)Ad`
C. `epsilon_(0)EAd`
D. `(1)/(2)epsilon_(0)E^(2)`

Explanation

To find the uniform electric field \( E \) between the plates of a parallel plate capacitor with area \( A \), plate separation \( d \), potential \( V_0 \), and dielectric constant \( K \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between electric field, potential, and distance**:
The electric field \( E \) between the plates of a capacitor is related to the potential difference \( V_0 \) and the distance \( d \) between the plates by the formula:
\[ E = \frac{V_0}{d} \]

2. **Consider the effect of the dielectric**:
When a dielectric material with dielectric constant \( K \) is present between the plates, the potential difference \( V_0 \) is modified. The relationship becomes:
\[ V_0 = E \cdot d \cdot K \]
Thus, we can express the electric field \( E \) in terms of \( V_0 \) and \( K \):
\[ E = \frac{V_0}{K \cdot d} \]

3. **Final expression for the electric field**:
Therefore, the uniform electric field \( E \) between the plates of the capacitor can be expressed as:
\[ E = \frac{V_0}{K \cdot d} \]

### Summary of the Solution:
The uniform electric field \( E \) between the plates of a parallel plate capacitor with dielectric constant \( K \) is given by:
\[ E = \frac{V_0}{K \cdot d} \]