5
$\begingroup$

Why is a dielectric slab repelled when inserted between the plates of a parallel plate capacitor which is connected to a battery of constant emf?

I think that it should get repelled as when the dielectric is inserted in the capacitor (connected to a battery) it increases the potential energy . And in most simple terms , the system tries to decrease its energy by repelling it . But I don't know if it is correct . And if it is correct please can you elaborate how it actually happens.

$\endgroup$
  • 4
    $\begingroup$ Why do you think the slab is repelled? $\endgroup$ – Ján Lalinský Aug 13 '16 at 14:45
  • 1
    $\begingroup$ Because my book says so , it also says that when there is no battery connected the dielectric will get attracted inside the capacitor . I can understand why it gets attracted as the charges on the capacitor plates will attract those charges that are induced in the dielectric . But I fail to understand why it gets repelled when the capacitor is connected to a battery $\endgroup$ – Varun Chandra Aug 13 '16 at 14:58
  • 3
    $\begingroup$ Could you post a link or the relevant section of the text from the book? It seems to me the author is wrong on this. $\endgroup$ – Ján Lalinský Aug 13 '16 at 15:05
  • 1
    $\begingroup$ But the video claims that slab is being pulled into the capacitor by electric forces, which is right. So if your book claims it is being pushed out of it, this is in contradiction to the video. It seems you would like to ask a different question though: why is the slab not being pushed out of the capacitor, since that would decrease the electrostatic energy of the capacitor? $\endgroup$ – Ján Lalinský Aug 13 '16 at 16:09
  • 1
    $\begingroup$ @LoopBack what you're suggesting is the opposite of what Varun Chandra described. Where did you get that idea? It is false, battery or no battery, if capacitor plates are charged, the slab is attracted to the plates. $\endgroup$ – Ján Lalinský Apr 27 at 10:54
4
$\begingroup$

Due to opposite charges induced on the faces of the dielectric due to the capacitor plates, the slab is attarcted. You can also see that as the dielectric goes in there is a constant rearrangement of charges, which requires work to be done. Who van do this work? Of course, thr electric field of the capacitor does thereby decreasing its own electric field by 1/k (k= dielectric const.) And this decreases the field energy finally. Edge effects also play a role here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.