00:01
To solve this question, we need to find the capacitance of the capacitor, which is somewhat like this.
00:10
This has thickness of l.
00:12
So capacitance here, this one, and this one, they are same.
00:19
There are two capacitors in series.
00:22
So the capacitor, capacitance here is given by twice a epsilon not divided by d.
00:32
Minus l since this distance is d minus l divided by same is the case with this so the net capacitance of these two capacitors is cn equals 1 over c plus 1 over c and if you solve it you get c n equals a epsilon not divided by d minus l let's also calculate the charge stored on the capacitor when it is connected to the battery q equals cv, or just cv is enough.
01:28
So that will be a epsilon not v divided by d minus l.
01:43
Now to solve the first part which says find the change in energy when the plate, is removed while the battery is connected.
02:00
That means the potential is the same.
02:02
So in this case, you can simply use dw is equal to half.
02:09
Since when you remove the plate, you're changing the capacitance.
02:12
So dc, v square.
02:17
So let's find dc.
02:18
Dc is initial capacitance minus the final capacitance.
02:28
From this, this is your initial capacitance.
02:32
Initial capacitance now is a epsilon not divided by d minus l final capacitance is a epsilon not divided by d since you remove the plate now this gives you a epsilon not l divided by l d minus l i'm sorry this should be d now, just substitute that to find the work done.
03:17
So the work done, in this case, is equal to v squared divided by 2, a epsilon 0, l, i'm sorry, again, d, d minus l.
03:40
Now, in the second part, we are supposed to consider that battery is disconnected.
03:47
That means the charge is constant...