00:01
In the given problem, the area of the square plates will be equal to square of the side of the plate.
00:10
Means square of 10 centimeter or we can say 10 .0 into 10 dash to power minus 2 square meter square.
00:23
Or it will come out to be 10 .5 minus 2 meter squared.
00:30
This is the area of the plates which are.
00:32
Are forming the capacitor.
00:35
Now, when we fill the space between the plates by two dielectric slabs, suppose this is the total gap between the plates, which is given as 1 .00 centimeter.
00:54
Now, this is the region which is filled with first dielectric slab, having dialectic constant k1, and it's a region.
01:06
Width is 1 by 5 centimeter.
01:12
This is 1 by 5 centimeter and here this will be the another dialectic slab having dialectic constant k2 whose width is 4 by 5 centimeter now this these two dialectics slab will convert this single capacitor into two capacitors which will be supposed to be connected in a series combination.
01:46
So for first dielectric distance between the plates or we can say the width of the dielectric d1 is equal to 1 by 5 centimetre or we can write it 0 .2 into 10 dashed bar minus 2 meter, the electric constant k1 is 20 .0.
02:18
So the capacitance of this capacitor, the first capacitor, will come out to be c1 is equal to epsilon not a into k1 by d1 as area of the plates will remain same for both of these capacitors in series.
02:36
Now plugging in the known values, for epsilon not, this is 8 .854 into 10 dash bar minus 12.
02:44
For area, this is 10 dash bar minus 2.
02:47
And for k1, this is 20 .0 divided by d1, which is 0 .2 into 10 dash bar minus 2 meter.
02:58
So finally, this capacitance of the first capacitor comes out to be 8 .8854 into 10 dash bar, minus 10 ferret...