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Question:On charging a parallel plate capacitor to potential V the spacing between the plates is halved and a dielectric medium of E(permittivity)=10 is introduced between the plates without disconnecting the dc source. Explain using suitable expressions how the (i) capacitance(ii)electric field(iii) energy density of the capacitor change.

I couldn't understand the solution of (ii) part of Problem 31 question.

Why they wrote"The potential V remains constant even after the introduction of dielectric medium"? But the dielectric medium does interfere in the potential difference of the capacitor. right? Is there any mathematical explanation for this?

Also why in that (ii) part of the question they used V=EL formula and not E= σ/ε where σ=surface chatge density and ε=permittivity

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    $\begingroup$ Can you repeat the question here? $\endgroup$ – garyp Feb 26 '16 at 11:40
  • $\begingroup$ I would guess you're supposed to assume the potential is maintained at $V$. For example the capacitor might be connected to a battery of voltage $V$. So the capacitance and therefore the charge will be changed but the voltage will not. The field strength is then obviously just $V$ divided by the plate spacing. $\endgroup$ – John Rennie Feb 26 '16 at 12:15
  • $\begingroup$ @garyp..yes look at the question now..Edit:I also fixed the blur in the picture $\endgroup$ – nishant_boro Feb 26 '16 at 12:29

Is there any mathematical explanation for this

Yes, the voltage across the capacitor is

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

Since, as is stated in your post, the dielectric is introduced without disconnecting the DC source, the source maintains the voltage $V$ constant by increasing the charge $Q$ as the capacitance $C$ increases.

If one were to introduce an ammeter in series with the source, one would see a charging current as the plates are brought closer together and while the dielectric is introduced.

Now, if the DC source were disconnected before any changes to the capacitor are made, the charge $Q$ is fixed since there is no circuit through which charge can flow. In that case, the voltage $V$ must change inversely with the capacitance changes.

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  • $\begingroup$ One more question...Why they used V=EL formula and not E= σ/ε(node) where σ=surface charge density and ε(node)=permittivity $\endgroup$ – nishant_boro Feb 26 '16 at 14:14

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