00:01
Hi, everybody.
00:01
So we need to find the charge on each capacitor and the potential difference across each one and the potential difference across a and b.
00:10
So before we begin, this is the series and we're just going to kind of solve what we need here.
00:19
Okay.
00:23
So we're solving the c1 and c2 right here series.
00:27
So for that, we have four times four, four plus four.
00:39
And this we have two new f.
00:43
Okay.
00:44
Now for the second one, we have c1 plus c3.
00:51
I'm doing this whole thing.
00:54
Okay.
00:55
And here we have two new.
00:59
F plus 4 new f, which equals 6 new f.
01:07
Okay.
01:09
Now we have to the third, the third one.
01:12
So here we have the second one, okay, times the c4.
01:18
Now we're doing, you know, the entire circuit here.
01:21
Okay.
01:22
And so we have c2 times, actually plus c4.
01:30
Okay.
01:31
And so we have six times four, six plus four, and we get two point four new.
01:41
Okay.
01:44
And now what we also can find is the charge of the system from a to b, okay, bolts in a to b.
01:53
And we get 2 .4 times 10 to the 6f times 28 volts, which gives us a total charge of 67.
02:08
0 .2 times 10 to negative 6c or muc if you want, okay? now we can do a, which was asking us, so a was asking us the capsence.
02:24
So it's going to be the c4, or which is going to be 67 .2 times 10 to negative 6c, okay? and because c2 is also is q2 as well.
02:51
So that also equals this number, okay? and now we have that.
03:07
And so we also need to find, let me bring this down a little bit.
03:14
So we have the volt 2 equals q2 over c2.
03:25
Okay, and we get 67 at times 10 to negative 6.
03:31
Divide 6 times 10 to negative 6, which gives us a volt of 11 .2 volts.
03:39
Okay...