The charge of a parallel plate capacitor is varying as q = q0 sin `omega`t. Then find the magnitude of displacement current through the capacitor. (Plate Area = A, separation of plates = d)

Options

A. `q_(0)cos (omega t)`
B. `q_(0)omega sin omegat`
C. `q_(0)omega cos omegat`
D. `(q_(0)Aomega)/d cos omegat`

Answer

To find the magnitude of the displacement current through a parallel plate capacitor with a varying charge given by \( q = q_0 \sin(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Displacement Current**:
The displacement current \( I_d \) is defined in the context of Maxwell's equations. It accounts for the changing electric field in regions where there is no conduction current. The formula for displacement current is given by:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux.

2. **Calculate the Electric Flux**:
The electric flux \( \Phi_E \) through the capacitor can be expressed as:
\[ \Phi_E = E \cdot A \]
where \( E \) is the electric field between the plates and \( A \) is the area of the plates. The electric field \( E \) can be related to the charge \( q \) on the capacitor:
\[ E = \frac{q}{\epsilon_0 A} \]
Substituting \( q = q_0 \sin(\omega t) \):
\[ E = \frac{q_0 \sin(\omega t)}{\epsilon_0 A} \]
Therefore, the electric flux becomes:
\[ \Phi_E = E \cdot A = \frac{q_0 \sin(\omega t)}{\epsilon_0 A} \cdot A = \frac{q_0 \sin(\omega t)}{\epsilon_0} \]

3. **Differentiate the Electric Flux**:
Now we differentiate the electric flux with respect to time \( t \):
\[ \frac{d\Phi_E}{dt} = \frac{d}{dt} \left( \frac{q_0 \sin(\omega t)}{\epsilon_0} \right) = \frac{q_0 \omega \cos(\omega t)}{\epsilon_0} \]

4. **Substitute into the Displacement Current Formula**:
Now substitute \( \frac{d\Phi_E}{dt} \) back into the displacement current equation:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \cdot \frac{q_0 \omega \cos(\omega t)}{\epsilon_0} \]
The \( \epsilon_0 \) cancels out:
\[ I_d = q_0 \omega \cos(\omega t) \]

5. **Final Result**:
Thus, the magnitude of the displacement current through the capacitor is:
\[ I_d = q_0 \omega \cos(\omega t) \]