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, we will analyze the situation step by step using the principles of energy conservation in the context of a parallel plate capacitor connected to a battery.

### Step-by-Step Solution:

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

2. **Identifying Work Done**:
- Let \( W \) be the work done by the external forces to separate the plates.
- Let \( W_B \) be the work done by the battery.
- Let \( \Delta U \) be the change in potential energy of the capacitor.
- Let \( H \) be the heat loss in the circuit.

3. **Applying Conservation of Energy**:
- According to the work-energy theorem, the net work done by all forces is equal to the change in kinetic energy. Since the plates are moved very slowly, the initial and final kinetic energies are both zero. Therefore, the net work done is zero:
\[ W + W_B - \Delta U - H = 0 \]

4. **Rearranging the Equation**:
- Rearranging the above equation gives us:
\[ W + W_B = \Delta U + H \]

5. **Considering Heat Loss**:
- Since the battery has zero internal resistance, there is no heat loss in the circuit. Thus, \( H = 0 \).
- Substituting \( H = 0 \) into the equation gives:
\[ W + W_B = \Delta U \]

6. **Final Relationship**:
- The final relationship we have derived is:
\[ W + W_B = \Delta U \]

### Conclusion:
The correct relationship for the given case is:
\[ W + W_B = \Delta U \]