I have just been introduced to the concept of central forces, and to the fact that they are per definition conservative forces. I have looked up several examples of central forces (gravity, electric, and spring), but they cover just about all the conservative forces I have ever heard about. Are there any conservative forces that are not central?
There must be, because otherwise there would not be any point in having a subcategory for central forces, yet I cannot find any examples anywhere.
If $\phi=-xy$, and ${F}=-\nabla \phi=y \hat{i}+x \hat{j}$ is conservative but not central
A constant force is conservative but not central.
For example: $\vec F=F \hat x$
You can check that the curl of this force is $0$, hence it is conservative. Its potential energy function in 3D space would just be $V(x,y,z)=-Fx+V_0$, where $V_0$ is some constant value.
An example of this is the approximation of gravity near the Earth's surface. In this regime the force is assumed to be constant, and we get the same form as above for the potential energy. Of course gravity on larger scales is a central force for, say, planets in orbit around a central star, which is why I gave the general form first.
Another example of a conservative, non-central force is one that is a superposition (sum) of two central forces.
Paul's answer is great. But I just found out an error in the background information you mentioned: central forces aren't necessarily conservative forces. I'm writing it down so you may have a clearer understanding of the logic relationship between a 'central' force and a 'conservative' force.
For example we may take $$\vec F = x \cdot \hat r$$ and with a bit of calculation we have$$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}= \frac{y^3+yz^2}{(x^2+y^2+z^2)^{3/2}}-\frac{-yx^2}{(x^2+y^2+z^2)^{3/2}}=\frac y r \neq 0$$
Since that's the $\hat z$ term in $\nabla \times \vec F$, we can tell that the curl is not zero, hence the force being nonconservative.
For a central force to be conservative, it must also be spherically symmetric, namely its magnitude must be a function of distance $r$ only. With that we can express $F$ as the gradient of some scalar $T$
$$\vec F = f(r)\cdot \hat r = \nabla T$$
with T being the indefinite integrationof $f(r)$
$$T = \int f(r)$$
Since the curl of a gradient is always zero,that gives us $\nabla \times \vec F = 0$, the proof of $F$ being conservative we're looking for.
The total potential of any distribution of sources of a conservative potential is itself conservative, but not necessarily central.
