Consider a uniform electric field in the `hat (z)` direction. The potential is a constant.

Options

A. in all space
B. for any x for a given z
C. for any y for a given z
D. on the x-y plane for given z

Explanation

To solve the problem, we need to analyze the behavior of electric potential in the presence of a uniform electric field directed in the \( \hat{z} \) direction.

### Step-by-Step Solution:

1. **Understanding the Electric Field**:
- A uniform electric field in the \( \hat{z} \) direction means that the electric field vector \( \mathbf{E} \) can be represented as \( \mathbf{E} = E \hat{z} \), where \( E \) is a constant magnitude.
- Electric field lines point in the direction of the field, indicating that the potential decreases in the direction of the field.

**Hint**: Remember that electric field lines indicate the direction of force on a positive charge.

2. **Electric Potential and Its Relation to Electric Field**:
- The electric potential \( V \) is related to the electric field by the equation:
\[ \mathbf{E} = -\nabla V \]
- This means that the potential decreases as you move in the direction of the electric field.

**Hint**: The gradient of the potential gives the electric field direction and magnitude.

3. **Equipotential Surfaces**:
- Equipotential surfaces are surfaces where the electric potential \( V \) is constant. For a uniform electric field in the \( \hat{z} \) direction, these surfaces are perpendicular to the electric field lines.
- Since the electric field is in the \( z \) direction, the equipotential surfaces will be horizontal planes at different heights \( z \).

**Hint**: Equipotential surfaces are always perpendicular to electric field lines.

4. **Analyzing the Options**:
- **Option 1**: "In all space" - This option is incorrect because the potential varies with height \( z \). Different planes at different \( z \) values will have different potentials.
- **Option 2**: "For any \( x \) for a given \( z \)" - This option is correct. For a fixed \( z \), the potential does not depend on \( x \) or \( y \) since it is constant on that horizontal plane.
- **Option 3**: "For any \( y \) for a given \( z \)" - This option is also correct for the same reason as Option 2.
- **Option 4**: "On the \( xy \) plane for a given \( z \)" - This option is correct as well since all points on the same \( z \) plane have the same potential.

**Hint**: Evaluate each option based on the behavior of potential in relation to the electric field direction.

5. **Conclusion**:
- The correct options are 2, 3, and 4. The potential is constant for any \( x \) and \( y \) at a fixed \( z \), and on the \( xy \) plane for a given \( z \).

### Final Answer:
- The correct options are: **2, 3, and 4**.