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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?

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  • \$\begingroup\$ I'm sure I've seen this answered here before. \$\endgroup\$
    – Hearth
    Jul 18 '20 at 11:14
  • \$\begingroup\$ please link the answer \$\endgroup\$
    – 666User666
    Jul 18 '20 at 11:16
  • \$\begingroup\$ Related: Make energy out of a capacitor in the sea and Assembling and Dis-assembling a Charged Capacitor. Do either of these answer your question? \$\endgroup\$
    – Dave Tweed
    Jul 18 '20 at 11:31
  • \$\begingroup\$ quite the question \$\endgroup\$
    – 666User666
    Jul 18 '20 at 11:32
  • \$\begingroup\$ The opposite experiment is also interesting: charge up capacitor plates with dielectric in-place....then pull one plate away (introducing air dielectric). Of course, the "pull" must be done with an insulated puller, so that charge remains constant. Voltage rises dramatically, perhaps resulting in a flash-over. \$\endgroup\$
    – glen_geek
    Jul 18 '20 at 11:39
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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.

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  • \$\begingroup\$ I think there could more details on the dieletric energy part.. like why does the capacitor want a dieletric to be inserted ? like I am quite baffled by the fact that dieletric actually sends capacitor into a lower energy state. Like making the medium less conducting is equivalent to lowering energy hm \$\endgroup\$
    – 666User666
    Jul 18 '20 at 11:53
  • \$\begingroup\$ @DDD4C4U try this site: opentextbc.ca/universityphysicsv2openstax/chapter/… - it gives a little more explanation. It's a fundamental property in electrostatics like why party balloons stick to a ceiling when rubbed - electrostatic forces are generated due to charge and attractions are made. You also said this that doesn't make any sense so I can't answer: Like making the medium less conducting is equivalent to lowering energy hm - conduction has nothing to do with it and I don't know what hm is. \$\endgroup\$
    – Andy aka
    Jul 18 '20 at 11:59
  • \$\begingroup\$ doesn't introducing a dieletric effect the conductivity of current between plates? ( displacement current) \$\endgroup\$
    – 666User666
    Jul 18 '20 at 12:04
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    \$\begingroup\$ The battery is disconnected so, there is no closed-circuit as such and, charge is conserved and there can be no current flow (displacement or normal). The energy is returned when the dielectric is removed hence no long-term energy loss. \$\endgroup\$
    – Andy aka
    Jul 18 '20 at 12:12
  • \$\begingroup\$ @DDD4C4U are we done with this Q and A now? \$\endgroup\$
    – Andy aka
    Jul 21 '20 at 12:47

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