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1. Jun 15, 2016

### RoboNerd

1. The problem statement, all variables and given/known data

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

2. Relevant equations
U = 0.5 * Q * deltaV

3. 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?

2. Jun 15, 2016

### harsh_sinha

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.

3. Jun 15, 2016

### Delta²

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?))

4. Jun 16, 2016

### RoboNerd

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?

5. Jun 16, 2016

### harsh_sinha

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.

6. Jun 17, 2016

### Delta²

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.

7. Jun 19, 2016

### RoboNerd

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

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