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 summetrically between the plates. Find the displacement current through this area.

Explanation

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

### Step 1: Understand the Concept of Displacement Current
Displacement current is defined as the rate of change of electric flux through a surface. It is given by the formula:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( I_d \) is the displacement current, \( \epsilon_0 \) is the permittivity of free space, and \( \Phi_E \) is the electric flux.

### Step 2: Determine the Electric Field Between the Plates
For a parallel-plate capacitor, the electric field \( E \) between the plates can be expressed as:
\[ E = \frac{Q}{A \epsilon_0} \]
where \( Q \) is the charge on the plates and \( A \) is the area of the plates.

### Step 3: Calculate the Electric Flux Through the Area A/2
The electric flux \( \Phi_E \) through a surface is given by:
\[ \Phi_E = E \cdot A \]
For the area \( A/2 \) that is parallel to the plates, the electric flux becomes:
\[ \Phi_E = E \cdot \frac{A}{2} = \frac{Q}{A \epsilon_0} \cdot \frac{A}{2} = \frac{Q}{2 \epsilon_0} \]

### Step 4: Relate Charge to Current
The current \( i \) is defined as the rate of change of charge:
\[ i = \frac{dQ}{dt} \]
Thus, we can express \( \frac{dQ}{dt} \) as \( i \).

### Step 5: Calculate the Rate of Change of Electric Flux
Now, we can find the rate of change of electric flux:
\[ \frac{d\Phi_E}{dt} = \frac{d}{dt} \left( \frac{Q}{2 \epsilon_0} \right) = \frac{1}{2 \epsilon_0} \frac{dQ}{dt} = \frac{i}{2 \epsilon_0} \]

### Step 6: Substitute into the Displacement Current Formula
Now we can substitute \( \frac{d\Phi_E}{dt} \) into the displacement current formula:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \cdot \frac{i}{2 \epsilon_0} = \frac{i}{2} \]

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

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