Charging a capacitor and then introducing a dielectric

Question

Suppose I have a capacitor and I charge it up using a circuit with a battery to the maximum amount of charge that it could hold.

Now, I unplug the wires and insert a dielectric inside.

I know that electrostatic energy between the capacitor plates changes when I do this, however, where does this energy go/come from? Very specifically how does insertion of dieletric change the energy? Is there any intuition behind this?

Secondly, how would the charges on the capacitor change at the point where I remove the battery and also the point where I insert the dielectric?

Finally how would I find the maximum amount of charge that I can allocate onto the capacitor?

The kind of answer I am looking for:

One which explains using electronic demonstration motivated by some mathematically derived theoretical results

Further, I'd also like to know what would be the effect of 'induced charges' on these processes and if we can ignore them or not. How does charge conservation explain induced charges?

Answer 1

I unplug the wires and insert a dielectric inside. I know that electrostatic energy between the capacitor plates change when I Do this , however, where does this energy go / come from.

• Charge remains the same,
• Capacitance increases by X and
• Voltage drops by X
• Energy (\$CV^2/2\$) drops by X

Hence energy "appears" to have been lost but, energy is used when the dielectric enters the plates; it is attracted inwards between the plates and that apparent energy loss (due to work done in moving the dielectric) is retrieved when it is removed (using some small force). The dielectric material is pulled in by the charge on the plates.

how would the charges on the capacitor change at the point where I remove the battery and also the point where I insert the dielectric?

They don't change; charge is conserved throughout unless the dielectric has conduction.

Also, finally how would I find the maximum amount of charge that I can allocate onto the capacitor?

Well, this is limited by the applied voltage so if you can estimate the breakdown voltage between the plates then you are 90% there.