Energy required to charge up a capacitor

[Maxwell1]

The energy required to charge a capacitor is:

W = ½ε_rCV²

From this we see that a capacitor with a linear dielectric in between its plates requires a bigger energy to charge up to a given potential. My question is: Should this be intuitive?
My teacher said that it is because the parts of the electric field is canceled off by the bound charges. I guess I can understand that since you then have to pull in more charge overall. However - won't these charges being pulled in also experience a weaker repulsion due to the canceled off charges?

 

[berkeman]

The energy required to charge a capacitor is:

W = ½ε_rCV²

From this we see that a capacitor with a linear dielectric in between its plates requires a bigger energy to charge up to a given potential. My question is: Should this be intuitive?
My teacher said that it is because the parts of the electric field is canceled off by the bound charges. I guess I can understand that since you then have to pull in more charge overall. However - won't these charges being pulled in also experience a weaker repulsion due to the canceled off charges?

Welcome to the PF.

Your equation does not look correct. See for example this thread:

https://www.physicsforums.com/showthread.php?t=170393

If you want to include non-linear dielectric effects, then it seems like you would write the capacitance as a function of voltage, like:

$$W = \frac{1}{2}C(V)V^2$$

And use an integral to calculate how much energe is stored in charging up the cap...