## anonymous 3 years ago work done to move a capacitor in parrallel a distance x

1. yrelhan4

*

2. anonymous

|dw:1364587708013:dw|

3. yrelhan4

I was just tagging myself.

4. anonymous

herre i mean to move it away from the other by applying a force $F_A$

5. anonymous

*

6. anonymous

work done = change in internal energy

7. anonymous

1) Energy in a capacitor $$U={1\over2}{Q^2\over C}$$

8. anonymous

F=qE

9. anonymous

2) what changes when you separate the plates? capacitance: $C={\epsilon_0kA\over d}$

10. anonymous

so, $U={1\over2}{Q^2d\over\epsilon_0kA}$

11. anonymous

W = final energy - initial energy

12. anonymous

remember that Q does not change in this case, but C and V do

13. anonymous

@Jonask you are increasing the separation between the plates, right?

14. anonymous

yes i was hoping for $dW=Fdx$

15. anonymous

i dont see x in you expressions above

16. anonymous
17. anonymous

i would like us to do it intergration way

18. anonymous
19. anonymous

you do not necessarily have to use the integration way. total work done is independent of path followed; depends on the initial and final states only!