00:01
So this is a classic capacitor problem.
00:02
We know that it's made from two plates that have an area of 0 .2 -0 meters.
00:14
Ooh, that looks funny.
00:20
Let me fix that.
00:23
Great.
00:23
And then we want to get, we know the distance is 0 .8 centimeters.
00:29
So d is equal to 0 .008 meters.
00:35
And the battery it's connected to is 120 volts.
00:43
And we wanna get the capacitance.
00:46
So capacitance, for that we wanna use capacitance is equal to epsilon knot a over d.
00:52
We could have used the voltage if we had known the charge.
00:56
So i'm just gonna calculate epsilon not times this a divided by that d and see what i get.
01:03
So 8 .5 times 2 divided by 0 .0 .8.
01:16
And then i got 2 .21 times 10 to the minus 10 ferrad.
01:28
And then b asks for the charge in each plate.
01:34
So for b, c is q over v.
01:39
So q is equal to cv.
01:41
So we just need to multiply v times the capacitance that we just calculated.
01:46
So we want to take this whole thing and multiply it by 120.
01:51
So i got 2 .2 .67, sorry, 6xx times 10 to the minus 8.
02:04
Did i say 7 or 8? 2 .66 times 10 to the minus 8.
02:14
Coolums.
02:15
And next we want to find probably the energy or something.
02:19
Let's see.
02:21
Electric field.
02:23
So electric field is voltage divided by distance.
02:30
So 120 divided by.
02:32
By 0 .008 meters.
02:36
Plugging that into a calculator, i get 15 ,000 volts per meter.
02:53
And for d, the energy is stored in the capacitor.
02:58
And now we have energy...