• RoboNerd
In summary, when a capacitor is fully charged by a battery and a dielectric is then inserted, the voltage will change because the battery is disconnected. However, there will be no flow of charge, so the charge will remain the same. The potential energy of the capacitor will also remain the same because the electric field between the plates will decrease, balancing out the increase in capacitance. Therefore, the only true statement is that the capacitance will increase.

## Homework Statement

A capacitor, is fully charged by a battery. The battery is disconnected and a dielectric is inserted into the capacitor. Which of the following is true?

I. Voltage will stay the same
II. Potential energy of the capacitor will increase
III. capacitance will increase

## Homework Equations

U = 0.5 * Q * deltaV

## The Attempt at a Solution

I know that III is true as dielectrics by definition increase capacitance. How can I eliminate the other two to get the final answer of III only?

Battery maintains constant voltage.Since it is removed voltage will change.
But there is no flow of charge.Using the relation between Charge ,Potential and capacitance and this information find change in voltage and potential energy.

Technically you don't need to "eliminate" the other choices as long as you are certain that a specific choice is correct in a multiple choice question.

But I) is wrong simply because inserting a dielectric changes the electric field E and it is V=Ed where d the (constant) distance between the plates.

II) you can answer it yourself, using the equation you wrote in OP and calculating how voltage will change (increase or decrease)? (Charge will remain the same (why?))

Delta² said:
II) you can answer it yourself, using the equation you wrote in OP and calculating how voltage will change (increase or decrease)? (Charge will remain the same (why?))

Charge will remain the same because the battery is not connected to the capacitor, so the battery can not pump out more charged particles in response to the insertion of the dielectric.

harsh_sinha said:
Battery maintains constant voltage.Since it is removed voltage will change.
But there is no flow of charge

Why does a battery maintain constant voltage? And, I take it that there is no flow of charge because there is no potential difference caused by having a battery hooked up to the circuit?

RoboNerd said:
Charge will remain the same because the battery is not connected to the capacitor, so the battery can not pump out more charged particles in response to the insertion of the dielectric.

Why does a battery maintain constant voltage? And, I take it that there is no flow of charge because there is no potential difference caused by having a battery hooked up to the circuit?
Battery is like a pumping device. It pumps charge.When connected to a capacitor it pumps charge till steady state is achieved.In steady state when there is no flow of charge we can easily prove that potential difference across the capacitor is constant using Kirchoff's Law.

I wonder what have you been taught about the dielectric constant k:

1) A capacitor with dielectric of dielectric constant k, has capacitance ##C_k=kC_0## where ##C_0## the Capacitance with vacuum between its plates.
or
2) The electric field between the plates of a capacitor with dielectric k is ##E_k=\frac{E_0}{k}## where ##E_0## the electric field with vacuum between the plates.

Delta² said:
I wonder what have you been taught about the dielectric constant k:

1) A capacitor with dielectric of dielectric constant k, has capacitance Ck=kC0Ck=kC0C_k=kC_0 where C0C0C_0 the Capacitance with vacuum between its plates.
or
2) The electric field between the plates of a capacitor with dielectric k is Ek=E0kEk=E0kE_k=\frac{E_0}{k} where E0E0E_0 the electric field with vacuum between the plates.

yes, I know how a dielectric works. thank you.

I see how I should solve this problem. Thanks a lot everyone!