How Is Work Calculated When Removing a Dielectric from a Charged Capacitor?

[clarr1]

A parallel-plate capacitor filled with dielectric K=3.4 is connected to a 100 V battery. After charging, the dielectric is removed and the battery remains connected. What would be the work required to remove the dielectric.I've done a problem similar to the one above, however, the battery was disconnected. Also, this problem doesn't give the area (A) or distance (d) of the capacitor, which would seem essential. But, I haven't gotten the problem correct after 20 or so attempts so what would I know?

What I do know is that the potential energy with the dielectric is:

.5(3.4((8.85E-12*A)/d))*(100^2)

and without the dielectric, the capacitance is: C=(8.85E-12)A/d

and the potential energy here is: .5(C)(V^2)=U

however, I know that when the dielectric is removed, charge flows back into the battery. Would this charge be the difference in charge on the plates before and after the dielectric is removed. And will this charge change the voltage of the battery. So, would i have to find the change in charge and add it to the constant charge of the battery to find the new voltage of the battery after the dielectric is removed? and would this charge go into the potential energy equation for after the dielectric is removed? and then would i subtract the two?Thanks

 

[clarr1]

So I know how to get the potential energy before the dielectric is removed. But when it is removed, charge goes from the plates back to the battery, making the charge with and without the dielectric equal. So really, my main question is this: does the charge traveling back into the battery change the voltage output of the battery which would be included in the second potential energy equation?