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I have a capacitor (C) connected to a battery of emf 100V. Hence potential drop across my capacitor is 100V. Now I introduce a dielectric between the plates of capacitor which increases the capacitance to C'. My book says that placing a dielectric inside the capacitor reduces the potential difference between the plates by reducing Electric field strength between them. But when finding the increase in energy due to dielectric, they used

(1/2)(C'-C)(100V)²

Why do we use 100V as potential drop, eventhough we know that dielectric reduced the potential difference between the plates?

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    $\begingroup$ Does the book say you disconnect the battery first? $\endgroup$ Nov 23, 2020 at 4:28
  • $\begingroup$ It didn't describe anything about reconnecting. They are just replacing the same capacitor with dielectric filled in it. But even then how can a capacitor have one potential difference inside and a different potential difference outside? My has no information about this contradiction $\endgroup$ Nov 23, 2020 at 4:49
  • $\begingroup$ @relayman357 I've changed the question a bit, maybe you can now understand what my problem is $\endgroup$ Nov 23, 2020 at 5:58

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Your book is talking about a capacitor that is changed to 100 volts and not connected to a battery. Inserting a dielectric increases the capacitance $C$. Since the charge $Q$ doesn’t change (it has no where to go) the voltage $V$ across the capacitor drops due to the relationship

$$C=\frac{Q}{V}$$

if you increase $C$ and $Q$ is constant $V$ has to be less.

The energy equation applies if you insert the dielectric while capacitor is connected to the battery. The voltage is fixed at the emf of the battery. Increasing the capacitance with the voltage fixed increases the charge on the capacitor per the above equation. The battery is what increases the charge.

Hope this helps .

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  • $\begingroup$ So the formula(E'=E/K) for electric field in the presence of dielectric isn't applicable in the case of dielectric capacitor connected to a constant voltage source, since the electric field is maintained same irrespective of the dielectric constant places inside it, am I right? $\endgroup$ Nov 23, 2020 at 6:39
  • $\begingroup$ Yes you are right because for a parallel plate capacitor the magnitude of the electric field is $E=V/d$. So if the voltage is fixed and the distance $d$ between the plates is fixed the electric field is fixed. Keep in mind that although the capacitance increases when inserting the dielectric the reason the field doesn’t decrease is because the battery increases the charge and increasing the charge maintains the field. $\endgroup$
    – Bob D
    Nov 23, 2020 at 7:16
  • $\begingroup$ Now it's clear. Thank you $\endgroup$ Nov 23, 2020 at 8:09

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