In a certain region of space, the electric potential is `V (x, y, z) = Axy - Bx^2 + Cy`, where A, B and C are positive constants.
(a) Calculate the x, y and z- components of the electric field.
(b) At which points is the electric field equal to zero?

Explanation

To solve the given problem, we will follow these steps:

### Part (a): Calculate the x, y, and z-components of the electric field.

1. **Understanding the relationship between electric potential and electric field**:
The electric field **E** is related to the electric potential **V** by the equation:
\[ \vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}\right) \]
where \(\nabla V\) is the gradient of the potential.

2. **Given potential**:
The electric potential is given as:
\[ V(x, y, z) = Axy - Bx^2 + Cy \]

3. **Calculate the partial derivatives**:
- **For \(E_x\)**:
\[ E_x = -\frac{\partial V}{\partial x} = -\left(\frac{\partial}{\partial x}(Axy - Bx^2 + Cy)\right) \]
\[ = -\left(Ay - 2Bx\right) = -Ay + 2Bx \]

- **For \(E_y\)**:
\[ E_y = -\frac{\partial V}{\partial y} = -\left(\frac{\partial}{\partial y}(Axy - Bx^2 + Cy)\right) \]
\[ = -\left(Ax + C\right) \]

- **For \(E_z\)**:
\[ E_z = -\frac{\partial V}{\partial z} = -\left(\frac{\partial}{\partial z}(Axy - Bx^2 + Cy)\right) \]
Since there is no \(z\) term in \(V\), we have:
\[ E_z = 0 \]

4. **Final expressions for the electric field components**:
Thus, we have:
\[ E_x = -Ay + 2Bx \]
\[ E_y = -Ax - C \]
\[ E_z = 0 \]

### Part (b): At which points is the electric field equal to zero?

1. **Setting the electric field components to zero**:
The electric field is zero when both \(E_x\) and \(E_y\) are zero:
\[ E_x = -Ay + 2Bx = 0 \quad (1) \]
\[ E_y = -Ax - C = 0 \quad (2) \]

2. **Solving equation (2)** for \(x\):
From equation (2):
\[ -Ax - C = 0 \implies Ax = -C \implies x = -\frac{C}{A} \]

3. **Substituting \(x\) into equation (1)**:
Substitute \(x = -\frac{C}{A}\) into equation (1):
\[ -Ay + 2B\left(-\frac{C}{A}\right) = 0 \]
\[ -Ay - \frac{2BC}{A} = 0 \implies Ay = -\frac{2BC}{A} \implies y = -\frac{2BC}{A^2} \]

4. **Conclusion**:
The electric field is zero at the points:
\[ \left(-\frac{C}{A}, -\frac{2BC}{A^2}, z\right) \quad \text{for any } z \]