A capacitor has been charged by a DC source. What are the magnitude of conduction and displacement current when it is fully charged?

Options

A. `I_(d) = I_(c) = 1`
B. `I_(d) = I_(c) = 0`
C. `I_(d) = 1, I_(c) = 0`
D. `I_(d) = 0, I_(c) = 1`

Explanation

To solve the problem, we need to analyze the behavior of a capacitor when it is fully charged by a DC source.

### Step-by-Step Solution:

1. **Understanding the Capacitor Charging Process**:
- When a capacitor is connected to a DC voltage source, it begins to charge. The current that flows through the circuit while charging is called the conduction current.

2. **Current Flow When Fully Charged**:
- Once the capacitor is fully charged, it reaches its maximum charge \( Q \). At this point, the flow of charge from the DC source stops because the voltage across the capacitor equals the voltage of the source. Therefore, the conduction current \( I_c \) becomes zero:
\[ I_c = 0 \quad \text{(when fully charged)} \]

3. **Displacement Current Concept**:
- The displacement current \( I_d \) is defined in the context of changing electric fields in a capacitor. It is given by the equation:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
- Here, \( \Phi_E \) is the electric flux, which depends on the electric field \( E \) and the area \( A \) of the capacitor plates.

4. **Electric Field in a Fully Charged Capacitor**:
- The electric field \( E \) between the plates of the capacitor when fully charged is given by:
\[ E = \frac{Q}{\epsilon_0 A} \]
- Since the charge \( Q \) is constant when the capacitor is fully charged, the electric field \( E \) is also constant.

5. **Electric Flux**:
- The electric flux \( \Phi_E \) is given by:
\[ \Phi_E = E \cdot A = \frac{Q}{\epsilon_0 A} \cdot A = \frac{Q}{\epsilon_0} \]
- Since \( Q \) is constant, the electric flux \( \Phi_E \) does not change over time.

6. **Displacement Current When Fully Charged**:
- Because the electric flux is constant, the rate of change of electric flux \( \frac{d\Phi_E}{dt} \) is zero:
\[ \frac{d\Phi_E}{dt} = 0 \]
- Therefore, the displacement current \( I_d \) also becomes zero:
\[ I_d = \epsilon_0 \cdot 0 = 0 \quad \text{(when fully charged)} \]

7. **Conclusion**:
- When the capacitor is fully charged, both the conduction current and the displacement current are zero:
\[ I_c = 0 \quad \text{and} \quad I_d = 0 \]

### Final Answer:
The magnitude of both conduction current and displacement current when the capacitor is fully charged is zero.

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