# Capacitor PE

1. Jan 23, 2009

### jan2905

The following 3 step process refers to a simple RC circuit with a battery and an initially open switch:

1- the switch is closed, allowing the capacitor to charge;
2- after the capacitor has charged, a slab of dielectric material is inserted between the plates of the capacitor and time passes;
3- the switch is opened, and the dielectric removed.

Q(capacitor)=CV(capacitor)
PE(capacitor)=1/2 CV^2

I said that: delta(PE)1>0; delta(PE)2>0; delta(PE)3<0

correct?

2. Jan 23, 2009

### LowlyPion

When you are putting the dielectric in the capacitor, you are effectively increasing the capacitance by the relationship that C/Co = ε the dielectric constant for the material, where ε > εo

As to your answers, I don't really know what the question is.

I would point out in step 3 that since the charge remains the same, but the capacitance is now influenced by εo instead of the higher ε of the removed dielectric, that the voltage will drop across the cap.

3. Jan 23, 2009

### jan2905

My question is whether or not the change in PE of the capacitor would reflect my "answer:" delta(PE)1>0; delta(PE)2>0; delta(PE)3<0 (the numerals are respective of the steps)

4. Jan 23, 2009

### LowlyPion

I guess I don't really understand your notation.

I just described what happens.

5. Jan 23, 2009

### epenguin

You decrease ε so you increase V according to pion's equation and physical intuition - you would need a higher voltage to establish that charge without the dielectric!?

6. Jan 23, 2009

### jan2905

yeah, that is what my answer shows right? a drop in V means a drop in PE. so PE final - PE initial < 0 ???

7. Jan 23, 2009

### LowlyPion

That's correct. I misstated it in typing it.

Thanks for catching that.

8. Jan 23, 2009

### jan2905

soooooo, i am correct?

9. Jan 23, 2009

### LowlyPion

No. I misstated my original description.

The equations were right. My English description was wrong.

10. Jan 23, 2009

### jan2905

ay.... i don't understand what happens then. this is my logic:

1- capacitor is charging (PE final is greater than PE initial therefore deltaPE>0)
2- capacitor has dielectric inserted (PE final is greater than PE initial because dielectric allows more capacitance therefore deltaPE>0)
3- capacitor no longer charging and dielectric is removed (PE final is less that PE initial because capacitor is now able to discharge therefore deltaPE<0)

11. Jan 24, 2009

### epenguin

3 does not correspond to 3 of your first post. Switch stays open, the charge has to stay there ideally. Yes, if the switch is then closed the capacitor will largely discharge.

12. Jan 24, 2009

### jan2905

yes 3 is the same on both posts:

3 (first post): the switch is opened, and the dielectric removed.
3 (last post): capacitor no longer charging [the switch is opened] and dielectric is removed

I am restating my logic. Is this correct? Anyone!

delta=final-initial; numerals correspond to the steps.

delta(PE)1>0; delta(PE)2>0; delta(PE)3<0