Work-energy theorem giving no work done on capacitor and dielectric when dielectric removed from capacitor For this problem,

Do you please know why when considering that the capacitor and dielectric is our system, the work energy theorem gives the wrong sign for the change in electrostatic potential energy
The correct solution is above,

From conservation of energy,

However from the work energy theorem,

 A: The work calculated in (b)(i) is the difference between the potential energy of the capacitor without and with the dielectric. This represents the minimum work required to transition from the initial to the final state.
This is analogous to doing work to roll a ball up hill, its (gravitational) potential energy increases while its kinetic energy can be kept constant. The work done by the force pushing the ball is equal and opposite to the work done by gravity. Here, the work done by electrostatic forces on the dielectric is equal and opposite to the work done by the force that removes it from the capacitor (i.e. what's given in (b)(i)).
A: If you use the work-energy theorem then you don't need to refer to potential energy. But you need to remember that the change in KE is equal to the net work. So, in your case, the work-energy theorem indeed "tells" that the net work is zero. This means that there are at least two forces doing work: the force pulling the slab from the capacitor and the force of the electric field on the slab. The work of the electric field is negative and in the critical case (minimum work done by the pulling force) is equal in magnitude to the positive work of the external, pulling force. This is why it is called the "minimum work". If the positive work is larger in magnitude than the work of the electric field then the net work is not zero and there will be some kinetic energy of the slab.
Edit
For the system capacitor+dielectric the energy is not really conserved because there are external forces. So if you want to use energy ballance (rather than "conservation") what we have is
$E_{final}=E_{initial} + W_{external}$
As the kinetic energy is zero in both states you simply have
$PE_{final}=PE_{initial} + W_{external}$
If you use the potential energy of the electric forces you have already counted for the work of these forces. For conservative forces (which are the ones with an associated PE) you either use the work itself or the potential energy but not both.
So, in the end, you have that the work done by that external force is
the difference between the potential energies:
$W_{external}=PE_{final}-PE_{initial} $
