# A dielectric is inserted between the plates of a capacitor.

1. Oct 2, 2007

### Xaspire88

A dielectric is inserted between the plates of a capacitor. The system is charged and the dielectric is removed. The electrostatic energy stored in the capacitor is

1. greater than
2. the same as
3. smaller than
it would have been if the dielectric were left in place.

U= 1/2 (K)$$E$$Ad E^2

K= dielectric constant

$$E$$= 8.85X10^-12

i would think that when the dielectric is removed that the electrostatic energy stored in the capacitor would be smaller than when the dielectric is in the capacitor assuming that K>1? but i have been informed that my thinking is incorrect. Any help would be appreciated.

2. Oct 2, 2007

### learningphysics

What is the energy stored in a capacitor in terms of charge and capacitance?

3. Oct 2, 2007

### Xaspire88

U=1/2 K((Q)^2)/C right?

4. Oct 2, 2007

why the K?

5. Oct 2, 2007

### Xaspire88

i was led to believe that the dielectric constant can be present in all of the energy equations for a capacitor. In this case we have a dielectric so wouldn't it be in the equation? It is removed but to see how it changes from when it was present it would have to be present in the original?

6. Oct 2, 2007

### learningphysics

The energy stored is $$\frac{Q^2}{2C}$$ regardless... C changes when the dielectric is removed... hence the energy changes.

7. Oct 2, 2007

### Xaspire88

And capacitance goes up when the dielectric is present meaning stored energy goes down. so when the dielectric is removed the capacitance down stored energy goes up?

8. Oct 2, 2007

yup.