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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Dielectric help needed!

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