00:01
Here we have a capacitor problem.
00:05
We are given the following configuration.
00:11
We have this parallel plate capacitor to be connected to some power supply, some voltage supply.
00:21
And the plates are square, let's say, of side lengths a.
00:28
And we are going to buy some potential here so that this capacitor will be hold some charge and we are asked to find this separation of the plates in order to provide these voltage and charge conditions.
00:53
So from a geometric point of view, the capacity is defined as the permittivity of the free space times the area of this plate divided by the separation of the plates and it does remember the value of the permit of the free space we have this value right over here and now from an electrostatic point of view we have this relation saying that the capacitance is equal to charged divided by the voltage applied so let us equate both sides.
01:52
Q over v equals epsilon 0, a or d.
01:57
Now we are given the q, v and a values in the following way.
02:06
For charge we have 12 .5 plus it's a number, not the area, but the a number to be inserted in terms of the of nanoclomes and for the voltage we have 34 .8 plus b volts that is equal to epsilon zero times so the side length is 6 .50 plus c centimeter and we are going to take the square to get the area divided by the separation.
02:53
Now let us solve this equation for d.
02:58
We get d equals epsilon 0 times 6 .50 plus c times 10 to the power minus 2 because i'm going to convert it to meters.
03:17
So this quantity squared times we have voltage value at 4 .a plus b volts divided by 12 .5 plus a times 10 to the power minus 9 columns.
03:42
So if we keep everything in the si units, we are going to get the d value in terms of meters.
03:48
So that's the beat of this argument.
03:51
Now we are given that this a value to be.
03:55
9, b to be 4 and c to be equal to 22.
04:04
If we insert these numbers into our calculators, we are going to get this the value to be 0 .00198 meters.
04:17
But we are asked to express this in millimeters and with three significant figures.
04:25
So we have 1 .0.
04:26
30 millimeters as the separation of these parallel plates.
04:35
Now there is the second sub -question here.
04:41
It says we are given some capacitors c1 and c2 they are connected in parallel and this whole combination is connected in series to this third capacitor.
05:06
And we are asked to find the equivalent capacitance in this circuit piece, let's say...
