# Capacitor with dielectric

there is a capacitor with a constant potential difference of V in air. When a dielectric is completely inserted between the plates of the capacitor completely filling it, additional charge flows onto the positive plate.

The additional charge should be Q=kCV-CV right?

Would the induced charge on either faces of the dielectric be Q=kCV?

Since the dielectric doesnt change the electric field (V=ED), why doesnt the increase in charge on the plates increase the field?

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learningphysics
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Yes, kCV is correct. extra charge is kCV - CV.

V = Ed

V is the same. d is the same. So E must be the same.

Yes, kCV is correct. extra charge is kCV - CV.

V = Ed

V is the same. d is the same. So E must be the same.

since there is an increase in charge on the plates, why doesnt this give rise to an increase in the E field since the increase in the chrage on the plates should serve to increase the field between the plates? I know V is the same and so E must be the same, I'm looking more for a conceptual idea.

also, there is kCV charge on the top on the dielectric right? does that mean that there is a -kCV charge on the bottom of the dielectric?

learningphysics
Homework Helper
since there is an increase in charge on the plates, why doesnt this give rise to an increase in the E field since the increase in the chrage on the plates should serve to increase the field between the plates? I know V is the same and so E must be the same, I'm looking more for a conceptual idea.

Well, the field is $$\frac{\sigma}{\epsilon}$$ which is $$\frac{Q}{A\epsilon}$$

so how does this change when the dielectric is inserted...

also, there is kCV charge on the top on the dielectric right? does that mean that there is a -kCV charge on the bottom of the dielectric?

yes.

charge increases when the dielectric is inserted so shouldnt the field increse too?

to: $$\frac{kCV}{A\epsilon}$$

learningphysics
Homework Helper
charge increases when the dielectric is inserted so shouldnt the field increse too?

to: $$\frac{kCV}{A\epsilon}$$

k is the dielectric constant. so $$\epsilon = k\epsilon_0$$

plug this in and the k's cancel in the numerator and denominator...

so you're left with

$$\frac{CV}{A\epsilon_0}$$, which is just like without the dielectric.

okay, so i want to understand this in a non mathematical way.

in the same situation as above, we insert a dielectric and then we get an increase in charge on the plates. the reason why we dont get an increase in the E-field is because the dielectric creates an opposing field that makes it so the E-field does not change. is the right?

learningphysics
Homework Helper
okay, so i want to understand this in a non mathematical way.

in the same situation as above, we insert a dielectric and then we get an increase in charge on the plates. the reason why we dont get an increase in the E-field is because the dielectric creates an opposing field that makes it so the E-field does not change. is the right?

Yes, that is correct.

but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?

learningphysics
Homework Helper
but since the dielectric has the same amount of charge as the metal plates, wouldnt the two E-fields cancel out?

No, just because the charge is the same doesn't mean the field is the same.

what is the field in the dielectric?

would it be:

$$E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}$$

learningphysics
Homework Helper
would it be:

$$E=\frac{kQ}{A\epsilon_o}-\frac{Q}{A\epsilon_o}$$

Yeah... I would write it as: $$E=\frac{Q}{A\epsilon_o}-\frac{kQ}{A\epsilon_o}$$

so when you add the field due to just the charge... you get the net field...