In a parallel plate capacitor, if the intervening medium of permittivity `epsilon` between the plates is replaced by another medium of permitivity `(epsilon)/(2)`, then its capacitance is

Options

A. halved
B. doubled
C. unchaned
D. quadrupled

Explanation

To solve the problem of how the capacitance of a parallel plate capacitor changes when the intervening medium is replaced, we can follow these steps:

### Step 1: Understand the formula for capacitance
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon A}{d} \]
where:
- \( \epsilon \) is the permittivity of the medium between the plates,
- \( A \) is the area of one of the plates,
- \( d \) is the separation between the plates.

### Step 2: Initial conditions
Let’s denote the initial permittivity as \( \epsilon \). Therefore, the initial capacitance \( C_1 \) can be expressed as:
\[ C_1 = \frac{\epsilon A}{d} \]

### Step 3: Change the medium
Now, the medium is changed to one with permittivity \( \frac{\epsilon}{2} \). We need to find the new capacitance \( C_2 \) with this new permittivity:
\[ C_2 = \frac{\frac{\epsilon}{2} A}{d} \]

### Step 4: Simplify the new capacitance
Simplifying \( C_2 \):
\[ C_2 = \frac{\epsilon A}{2d} \]

### Step 5: Compare the two capacitances
Now, we can compare \( C_1 \) and \( C_2 \):
- \( C_1 = \frac{\epsilon A}{d} \)
- \( C_2 = \frac{\epsilon A}{2d} \)

From this, we can see that:
\[ C_2 = \frac{1}{2} C_1 \]

### Conclusion
Thus, when the permittivity of the medium is halved, the capacitance of the capacitor is also halved. Therefore, the final answer is:
\[ C_2 = \frac{C_1}{2} \]

### Final Answer
The capacitance of the capacitor after replacing the medium is half of the original capacitance.

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