- #1
casanova2528
- 52
- 0
I'm not too fond of writing the math down here, so I need only those who know their circuits to reply.
Battery connection is maintained with one capacitor.
du goes up as you insert a dielectric
dc goes up
dq goes up
However, foe those who've done the thinking and solution...we find that dE of the system is negative!
The dielectric is basically sucked in if there is no external force opposing the attraction! That is very counter intuitive because we would assume that the increase in du as the dielectric is inserted would come only from the external force that would oppose the insertion!
How can you explain this truth about the sucking in of the dielectric? After all, charge flows from the battery onto the plates as the dielectric is inserted.
The mathematical answer lies in this part of the step to find the -dE answer.
dq (v) = dU battery. As opposed to using
.5 dq (v) = du battery.
I'm not sure how to explain this. Help me !
Battery connection is maintained with one capacitor.
du goes up as you insert a dielectric
dc goes up
dq goes up
However, foe those who've done the thinking and solution...we find that dE of the system is negative!
The dielectric is basically sucked in if there is no external force opposing the attraction! That is very counter intuitive because we would assume that the increase in du as the dielectric is inserted would come only from the external force that would oppose the insertion!
How can you explain this truth about the sucking in of the dielectric? After all, charge flows from the battery onto the plates as the dielectric is inserted.
The mathematical answer lies in this part of the step to find the -dE answer.
dq (v) = dU battery. As opposed to using
.5 dq (v) = du battery.
I'm not sure how to explain this. Help me !