Recent content by Jacob White

Ok, thank you for explanation!

Ok, but if didn't assume that they are very large what would happen? Do you have any idea how we could deal with situation like this?

Yes, in the case of infinite plates definitely moving charge will not affect induced charges. However originally they stated that the area of plate is A so it's finite.

I don't know why moving the point charge around, keeping it a constant distance from the plates keeps the propotion of induced charges at plates. It looks like it would be that way but I wonder if we can prove that somehow.

There is even another thread on PF on this problem where they suggested method of images however it is much more complicated and it is definitely not they had in mind.

I don't understand why moving charge doesn't affect the magnitude of induced charge. Could someone explain why it is possible? Thanks in advance.

Thanks for realizing me that it works only when we take heat only from 2 reservoirs!

I have said that too briefly. I have seen proof like this: If we had two heat reservoirs at T1 and T2 and two reversible engines A and B. So suppose A takes heat Q1 from T1 and doing work W1. B is also reversible so we can reverse it's cycle so it would use work W1 to transfer heat from T2 to...

Ok, now I see that indeed reversible engines would have the Carnot efficiency only when working between 2 heat reservoirs at given temperatures. And with different temperatures this argument of reversing cycle and using to produce additional work just doesn't work.

So we know that every reversible engine working between the same temperatures will have the same efficiency(the same as Carnot engine). So let's consider for example reversible Otto cycle. So as you can see on the picture it is operating between ##T_1## and ##T_3##, so I was thinking that it...

Sorry again, I forgot about molar masses. 3.737*18/29=2.32

0.5084 kPa

Much more - about 3.737g.

So now after these corrections the final temperature I get is 270.89K which is about 0.6K higher than without considering ice. And that makes sense since we get some additional energy from freezing water.

So at 73.535kPa there would be no water molecules left. So from this pressure we could just go back to our first numerical solution and simply change the condensation heat to sublimation heat?