If the electric field is not a gradient, can it exist? We know that the gradient of the electric potential function $V(x,y,z)$ is the electric field. But not all vector fields are gradients, for example $y\hat{i}-x\hat{j}$ is not a gradient. Does this mean that an electric field given by $\vec{E}=y\hat{i}-x\hat{j}$ cannot exist? In particular does this mean there is no arrangement of charges which could create such a field? What is the correct interpretation here?
Note: the example I have given here is a bit stupid, because the field is two dimensional and this is probably impossible anyways. But there are of course three dimensional fields which are not gradients, I just can't think of one at the moment. If someone can please let me know.

Edit: to answer the comment below I will show that $y\hat{i}-x\hat{j}$ is indeed not a gradient. If it is, then there exists a function
$f$ such that
$∇f(x, y) = yi − xj$.
Obviously
$\frac{∂f}{∂x}(x, y) = y$,
$\frac{∂f}{∂y}(x, y) = −x$ so
$\frac{∂^2f}{∂y∂x}(x, y) = 1$,
$\frac{∂f}{∂x∂y} (x, y) = −1$
and thus $\frac{∂^2f}{∂y∂x}(x, y) \neq \frac{∂f}{∂x∂y} (x, y)$.
This contradicts the mixed partial derivative theorem: the four partial derivatives
under consideration are everywhere continuous and thus, according to the mixed partial
derivative theorem, we must have
$\frac{∂^2f}{∂y∂x}(x, y) = \frac{∂f}{∂x∂y}(x,y)$.
 A: In electrostatics, there is always the scalar potential (voltage) such that $\vec\nabla\phi = -\vec E$.
This does not hold true in general electromagnetism, which is described by the electromagnetic four potential $A$ consisting of the electrostatic potential $\phi$ and the magnetic vector potential $\vec A$ such that
$$ \vec E = -\vec\nabla\phi - \partial_t \vec A$$
$$ \vec B = \vec\nabla\times\vec A$$
So, not being a gradient is not sufficient to conclude that a given vector field cannot be the electric field, since for non-zero $\vec A$, i.e. in the presence of currents/moving charges/magnetic fields, we do not expect $\vec E$ to be the gradient of a scalar field.
An electric field would be impossible if it cannot arise as the solution to Maxwell's equations, but as long as your magnetic field is unconstrained, you can obviously just set $\partial_t \vec B$ and $\rho$ such that the given field fulfills them, so you cannot from the form of $\vec E$ alone judge whether it is physically permissible.
