A parallel- plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Options

A. `I`
B. `(I)/(2)`
C. `(I)/(4)`
D. `(I)/(8)`

Explanation

To solve the problem of finding the displacement current through a plane surface of area A/2 parallel to the plates of a parallel-plate capacitor, we can follow these steps:

### Step 1: Understand the given parameters
We have a parallel-plate capacitor with:
- Plate area \( A \)
- Separation between the plates \( d \)
- Constant charging current \( i \)
- A plane surface of area \( A/2 \) drawn symmetrically between the plates.

### Step 2: Calculate the electric field between the plates
The electric field \( E \) between the plates of a parallel-plate capacitor is given by the formula:
\[ E = \frac{Q}{\epsilon_0 A} \]
where \( Q \) is the charge on the plates and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Determine the charge \( Q \) on the plates
Since the capacitor is charged by a constant current \( i \), the charge \( Q \) on the plates at any time \( t \) can be expressed as:
\[ Q = i \cdot t \]

### Step 4: Calculate the electric flux \( \Phi \) through the area \( A/2 \)
The electric flux \( \Phi \) through the area \( A/2 \) is given by:
\[ \Phi = E \cdot \frac{A}{2} \]
Substituting the expression for \( E \):
\[ \Phi = \left(\frac{Q}{\epsilon_0 A}\right) \cdot \frac{A}{2} = \frac{Q}{2\epsilon_0} \]

### Step 5: Find the displacement current \( I_d \)
The displacement current \( I_d \) is defined as:
\[ I_d = \epsilon_0 \frac{d\Phi}{dt} \]
Substituting the expression for \( \Phi \):
\[ I_d = \epsilon_0 \frac{d}{dt} \left(\frac{Q}{2\epsilon_0}\right) = \epsilon_0 \cdot \frac{1}{2\epsilon_0} \frac{dQ}{dt} \]
This simplifies to:
\[ I_d = \frac{1}{2} \frac{dQ}{dt} \]

### Step 6: Substitute the constant current \( i \) for \( \frac{dQ}{dt} \)
Since \( \frac{dQ}{dt} = i \), we have:
\[ I_d = \frac{1}{2} i \]

### Final Answer
Thus, the displacement current through the area \( A/2 \) is:
\[ I_d = \frac{i}{2} \]

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