The electric potential at a point (x, y, z) is given by V = -x²y - xz³ + 4. The electric field E at that point

Question

The electric potential at a point (x, y, z) is given by V = -x²y - xz³ + 4. The electric field E at that point is

Accepted Answer

1. **Relationship between Field and Potential:** The electric field $\mathbf{E}$ is related to the electric potential $V$ by the negative gradient: $\mathbf{E} = -
abla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k})$ [1].
2. **Partial Derivative with respect to x:**
   $\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3$
   $E_x = -(-2xy - z^3) = 2xy + z^3$
3. **Partial Derivative with respect to y:**
   $\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2$
   $E_y = -(-x^2) = x^2$
4. **Partial Derivative with respect to z:**
   $\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2$
   $E_z = -(-3xz^2) = 3xz^2$
5. **Resulting Vector:**
   $\mathbf{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$

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