1
$\begingroup$

Under what conditions does the field due to free charge is simply reduced by factor of $1/k$ where $k$ is dielectric constant? I.e my book says if dielectric is entirely filled in space.If it is so, how can we apply it between a capacitor? Because dielectric is just between capacitor not the entire space.

$\endgroup$

1 Answer 1

1
$\begingroup$

The dielectric has to be in the space where the electric field is.

In an ideal capacitor the field is only present between the two plates. So you can put in an dielectric to increase the capacity of the capacitor. The outside space doesn't matter because there's no relevant field that could be affected by a change of the dielectric constant.

$\endgroup$
7
  • $\begingroup$ So we can't apply it if capacitor if half filled with dielectric? $\endgroup$ Commented Jul 28, 2016 at 8:14
  • 1
    $\begingroup$ Of course you can. If the capacitor is half filled, so one part is empty and the other is full you can say that it equals two capacitors in parallel. One with dielectric and the other one without. If the dielectric changes along the electrical field. Lets say you have a capacitor with a plate distance of 1 cm and you have a dielectric on one plate with 0.5 cm thickness. The simple way is to assume you have two capacitors in series one with the dielectric and the other one without. $\endgroup$
    – GNA
    Commented Jul 28, 2016 at 8:20
  • $\begingroup$ Thanks! But I have confusion in half filled ( 2nd one u described) how can we assume that capcitor as a series of two capacitor. Because an ordinary capacitor has equal and opposite charge on opposite plates. $\endgroup$ Commented Jul 28, 2016 at 8:35
  • $\begingroup$ Look at this picture. It shows the relation of a series of capacitors to a single one. The metal plates and the wire in the middle do not carry any electrical field. So two capacitors in series with d1 and d2 as plate distance are equal to one cap with a plate distance of d1+d2 . This is also shown here $\endgroup$
    – GNA
    Commented Jul 28, 2016 at 8:40
  • $\begingroup$ And why don't we can apply it ( for the field inside dielectric) also when dielectric is not present at some space but electric field is? $\endgroup$ Commented Jul 28, 2016 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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