Could someone explain how they arrive at $-\nabla V=-\frac{P}{3\epsilon_0}\hat z$? If I use the gradient in polar coordinates, I get:
$$
\nabla V=\frac{P}{3\epsilon_0}\cos\theta\,\hat r-\frac{P}{3\epsilon_0}\sin\theta\,\hat\theta.
$$
I’m guessing $\hat z$ is just the unit vector that points in the positive $z$ direction. So somehow we would need: $\cos\theta\,\hat r -\sin\theta\,\hat\theta=\hat z$.