If the gap between the plates of a parallel plate capacitor is filled with medium of dielectric constant k=2 , then the field between them

Options

A. Increases by a factor 2
B. Increases by a factor `sqrt(2)`
C. Decreased by a factor `sqrt(2)`
D. Decreased by a factor `1//2`

Explanation

To solve the problem of finding the electric field between the plates of a parallel plate capacitor when a dielectric medium with a dielectric constant \( k = 2 \) is introduced, we can follow these steps:

### Step 1: Understand the Initial Condition
Initially, we have a parallel plate capacitor with air (or vacuum) between the plates. The electric field \( E \) in this case can be expressed as:
\[ E = \frac{Q}{A \epsilon_0} \]
where \( Q \) is the charge on one plate, \( A \) is the area of the plates, and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Introduce the Dielectric Medium
When a dielectric medium with dielectric constant \( k \) is introduced between the plates, the electric field \( E' \) in the presence of the dielectric is given by:
\[ E' = \frac{E}{k} \]
This means the electric field is reduced by a factor of \( k \).

### Step 3: Substitute the Dielectric Constant
Given that the dielectric constant \( k = 2 \), we can substitute this value into the equation:
\[ E' = \frac{E}{2} \]

### Step 4: Relate the New Electric Field to the Original Electric Field
Since we already established that the original electric field \( E \) is:
\[ E = \frac{Q}{A \epsilon_0} \]
We can express the new electric field as:
\[ E' = \frac{1}{2} \cdot \frac{Q}{A \epsilon_0} \]

### Step 5: Conclusion
Thus, the electric field between the plates when the dielectric medium is introduced is reduced by a factor of 2:
\[ E' = \frac{E}{2} \]

### Final Answer
The electric field between the plates is decreased by a factor of 2.

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