When a capacitor is charged then
`l_(c)`= conduction current `l_(d)` = displacembet current)

Options

A. `I_c = 0, I_d != 0`
B. `I_c != 0, I_d = 0`
C. `I_c = I_d `in same direction
D. `I_c = I_d in opposite direction`

Answer

To solve the problem regarding the relationship between conduction current \( I_c \) and displacement current \( I_d \) when a capacitor is charged, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept**: When a capacitor is charged, there are two types of currents involved: conduction current \( I_c \) and displacement current \( I_d \). The conduction current is the actual flow of charge through the conductor, while the displacement current is related to the changing electric field in the capacitor.

2. **Expression for Conduction Current**: The conduction current \( I_c \) can be expressed in terms of the charge \( Q_t \) on the capacitor:
\[ I_c = \frac{dQ_t}{dt} \]
This equation states that the conduction current is the rate of change of charge with respect to time.

3. **Expression for Displacement Current**: The displacement current \( I_d \) is defined in terms of the electric flux \( \Phi_E \):
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \epsilon_0 \) is the permittivity of free space.

4. **Calculating Electric Flux**: The electric flux \( \Phi_E \) due to the electric field \( E \) inside the capacitor can be expressed as:
\[ \Phi_E = E \cdot A \]
where \( A \) is the area of the capacitor plates. The electric field \( E \) can be expressed in terms of charge density \( \sigma \):
\[ E = \frac{\sigma}{\epsilon_0} = \frac{Q_t/A}{\epsilon_0} = \frac{Q_t}{\epsilon_0 A} \]
Therefore, the electric flux becomes:
\[ \Phi_E = \frac{Q_t}{\epsilon_0} \cdot A \]

5. **Substituting into Displacement Current**: We can substitute the expression for electric flux into the displacement current equation:
\[ I_d = \epsilon_0 \frac{d}{dt}\left(\frac{Q_t}{\epsilon_0}\right) = \frac{dQ_t}{dt} \]
Here, \( \epsilon_0 \) cancels out, leading to:
\[ I_d = \frac{dQ_t}{dt} \]

6. **Comparing the Two Currents**: From the above equations, we see that:
\[ I_c = \frac{dQ_t}{dt} \quad \text{and} \quad I_d = \frac{dQ_t}{dt} \]
Thus, we conclude that:
\[ I_c = I_d \]

7. **Direction of Currents**: Since both currents are equal and arise from the same physical process, they flow in the same direction.

8. **Final Conclusion**: Therefore, the correct statement is:
\[ I_c = I_d \quad \text{(both have the same direction)} \]