00:01
Hello students, in this question we are given that initially two large metallic parallel plates each of area a are separated at a distance of 3d and they are connected to a grounded wire which joins them and initially we are given that they are uncharged.
00:18
Now and another plate which is carrying a total charge of q is placed at a distance of d from the upper plate and we have to calculate the first we have to calculate the induce charges on the plate.
00:30
So as we can see that initially we were given these two plates that are named 1 and 2 and later on another plate which is named as 3 is placing between these plates at a distance d from the upper plate which is carrying a charge of q.
00:50
We are given that each plate has an area of a.
00:54
So first let us see that the upper plate and the middle plate of this arrangement, forms and capacitor which we name as c1 and we can see that this is c1 and we can also see that the middle and the lower plate forms in another capacitor which can be named as c2.
01:13
Now if we calculate the capacitance of this capacitor so capacitance of c1, capersitence of c1 is equal to epsilon not a divided by d as the.
01:31
The area between the plates is d and each place has an area of a, so the capacitance c1 is equal to epsilon knot multiplied by a divided by d and similarly we can write that capacitance of capacitor c2 that would be equal to the epsilon not which is the permutivity of free face multiplied by the area of the plates and divided by the distance between them which is given to us as 2d.
02:00
As the initially the distance between the plate is 3d and the distance between the upper and the middle plate is d, so the distance between the middle and the lower plate would be equal to 2e.
02:12
So this is the capacitance of c1 and c2.
02:15
And upon calculating the c total for this arrangement that would be equal to the c net as they are placed in parallel so that would be equal to c1 plus c2...