The separation between the plates of a parallel plate capacitor , connected to a battery (zero resistance) of constant EMF is increased with constant (very slow) speed by external forces . During the process, w is the work done dy external forces. `DeltaU` is the change in potential energy of the capacitor , `w_b` is work done by the battery and H is the heat loss in the circuit . Then

Options

A. `w+w_b=DeltaU`
B. `Hne0`
C. `H=DeltaU`
D. `w=0`

Answer

To solve the problem regarding the parallel plate capacitor connected to a battery, we will analyze the work done by external forces, the work done by the battery, the change in potential energy, and the heat loss in the circuit.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a parallel plate capacitor connected to a battery with a constant EMF (E).
- The separation between the plates of the capacitor is increased very slowly by external forces.

2. **Work Done by External Forces (W)**:
- As the separation between the plates increases, external forces do work on the system. We denote this work as \( W \).

3. **Work Done by the Battery (W_b)**:
- The battery maintains a constant EMF, which means it does work to keep the potential difference across the capacitor constant. We denote this work as \( W_b \).

4. **Change in Potential Energy (ΔU)**:
- The potential energy stored in a capacitor is given by \( U = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.
- As the capacitor's separation increases, its capacitance \( C \) decreases, and thus the potential energy changes. We denote this change as \( \Delta U \).

5. **Heat Loss in the Circuit (H)**:
- Since the battery is ideal (zero resistance), there is no heat loss in the circuit. Therefore, \( H = 0 \).

6. **Applying the Work-Energy Principle**:
- According to the work-energy principle, the total work done on the system is equal to the change in internal energy. In this case, the internal energy change is represented by the change in potential energy \( \Delta U \).
- Therefore, we can write the equation:
\[ W + W_b = \Delta U \]

7. **Conclusion**:
- Since there is no change in kinetic energy (the process is very slow), the work done by the external forces and the work done by the battery equals the change in potential energy of the capacitor.
- Thus, the correct relationship is:
\[ W + W_b = \Delta U \]
- Since \( H = 0 \), there is no heat loss in the circuit.

### Final Result:
The relationship between the work done by external forces, work done by the battery, and the change in potential energy of the capacitor is given by:
\[ W + W_b = \Delta U \]