00:01
All right, so in this problem, we have the capacity.
00:07
And initially it is filled by air.
00:09
So we have the electric equal 1.
00:13
So, and then we insert this with myla, and the electric constant of myla is 3 .1.
00:24
And we connected the two plates with constant voltage, so v is 12 volts.
00:31
And we also know that before, before we insert this mylar material, the capacity is 0 .25 microferrales.
00:42
So in part a, we want to figure out the change of the charge on the two plates.
00:50
So before we insert the myla, q equals c times v, right? so c is right here, and v is here.
00:58
So see that the value is 3 times 10 to negative 6 columns.
01:02
And after we insert myla q prime equal c prime times v, right? because v is constant.
01:11
And c prime c is proportion to c has expression kappa a epsilon n a over d, right? so c is a proportional to kappa.
01:21
So this is a so q prime in this sense is proportional to kappa.
01:25
So this equal 9 .3 times 10 to negative 6 columns...