Can the expression E = xyî + 2yzĵ + 3xz k be said to be that of an electrostatic field? Justify your answer.

The curl operator is given by:

$$\nabla \times \vec{E} = \left( \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} \right)\hat{i} + \left( \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right)\hat{j} + \left( \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right)\hat{k}$$

For the $\hat{i}$ component:

$$\frac{\partial E_z}{\partial y} = \frac{\partial}{\partial y}(3xz) = 0$$ $$\frac{\partial E_y}{\partial z} = \frac{\partial}{\partial z}(2yz) = 2y$$

So, the $\hat{i}$ component is $0 - 2y = -2y$.

For the $\hat{j}$ component:

$$\frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z}(xy) = 0$$ $$\frac{\partial E_z}{\partial x} = \frac{\partial}{\partial x}(3xz) = 3z$$

So, the $\hat{j}$ component is $0 - 3z = -3z$.

For the $\hat{k}$ component:

$$\frac{\partial E_y}{\partial x} = \frac{\partial}{\partial x}(2yz) = 0$$ $$\frac{\partial E_x}{\partial y} = \frac{\partial}{\partial y}(xy) = x$$

So, the $\hat{k}$ component is $0 - x = -x$.

Since $\nabla \times \vec{E} = -2y\hat{i} - 3z\hat{j} - x\hat{k}$ is not equal to zero, the given expression for $\vec{E}$ does not represent an electrostatic field. An electrostatic field must have a curl of zero because it is a conservative field, meaning the line integral of the electric field around any closed loop is zero.

The expression below cannot be said to be that of an electrostatic field.
Justification: An electrostatic field is a conservative field, which implies that its curl must be zero. The calculated curl of the given field is $\nabla \times \vec{E} = -2y\hat{i} - 3z\hat{j} - x\hat{k}$, which is not zero.

The final answer is $\boxed{\text{No, the expression does not represent an electrostatic field.}}$