The charge on a parallel plate capacitor varies as `q = q_(0) "cos" 2 pit`. The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor is

Options

A. `q_(0)2pi upsilon sin pi upsilont`
B. `-q_(0)2pi upsilon sin 2pi upsilont`
C. `q_(0)2pi sin pi upsilont`
D. `q_(0)pi upsilon sin2pi upsilont`

Explanation

To find the displacement current through a parallel plate capacitor where the charge varies as \( q = q_0 \cos(2 \pi t) \), we will follow these steps:

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

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

### Step 3: Calculate the electric flux
The electric flux \( \Phi \) through the capacitor is given by:
\[ \Phi = A \cdot E = A \cdot \frac{Q}{A \epsilon_0} = \frac{Q}{\epsilon_0} \]

### Step 4: Substitute the expression for charge
Given that \( Q = q_0 \cos(2 \pi t) \), we can substitute this into the expression for electric flux:
\[ \Phi = \frac{q_0 \cos(2 \pi t)}{\epsilon_0} \]

### Step 5: Differentiate the electric flux
Now, we need to find the rate of change of electric flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt} \left( \frac{q_0 \cos(2 \pi t)}{\epsilon_0} \right) = \frac{q_0}{\epsilon_0} \frac{d}{dt} \left( \cos(2 \pi t) \right) \]
Using the derivative of cosine, we have:
\[ \frac{d}{dt} \left( \cos(2 \pi t) \right) = -2 \pi \sin(2 \pi t) \]
Thus,
\[ \frac{d\Phi}{dt} = \frac{q_0}{\epsilon_0} (-2 \pi \sin(2 \pi t)) = -\frac{2 \pi q_0 \sin(2 \pi t)}{\epsilon_0} \]

### Step 6: Substitute back to find displacement current
Now, substituting this back into the formula for displacement current:
\[ I_d = \epsilon_0 \frac{d\Phi}{dt} = \epsilon_0 \left( -\frac{2 \pi q_0 \sin(2 \pi t)}{\epsilon_0} \right) \]
The \( \epsilon_0 \) cancels out:
\[ I_d = -2 \pi q_0 \sin(2 \pi t) \]

### Final Answer
The displacement current through the capacitor is:
\[ I_d = -2 \pi q_0 \sin(2 \pi t) \]