00:01
Hi, here in this given problem capacitance of the parallel plate capacitor that is given to be equal to 3 .50 micro ferret.
00:35
In the first part of the problem, potential applied across its plates that is 21 .0 volt.
00:51
So energy stored in this capacitor will be given by the expression half c.
01:06
V squared.
01:08
Plugging in the known values, this is half times of capacitance 3 .50 micro ferret multiplied by square of 21.
01:23
So this energy stored in the capacitor is calculated to be equal to 772 microjoules.
01:36
Answer for the first part of this given problem here.
01:43
In the second part of the problem, battery is disconnected.
02:02
So the charge over the plates of the capacitor that will remain same.
02:25
So first of all we find the charge is stored over the capacitor which is given by the expression q is equal to c multiplied by v means this is 3 .50 micro ferret excuse me micro ferret multiplied by 21 volt so this charge is stored over the capacitor is calculated to be equal to 73 .5 micro ferrette.
02:59
Croculum and as the distance between the plates, plates of the capacitor has been reduced to half.
03:29
So using an expression for the capacitance of parallel plate capacitor, c is equal to epsilon not a divided by d...