00:01
In this question we have basically three parts that we have the calculates.
00:03
That part is known as part b in the question part d and part f.
00:08
Okay, so we will begin by writing the given data.
00:11
Firstly, v is given to us 9 volts.
00:16
D is given to us that is 0 .10 millimeter in terms of meter rate will be 0 .10 times stainless to the power minus 3 meter.
00:34
So now, let's solve for part b that is given to us in the that will be a capacitor electric field.
00:47
So we know that e is equals to v over d, that is equal to, v is equal to 9 over d, that is equals to 0 .10 times 10 to the power minus 3.
01:09
This is equal to 9 times 10 less to the power 4, volt per meter.
01:18
This will be the answer, okay, for part b.
01:27
Now part d.
01:32
Part d is a bit theoretical.
01:33
You have to just understand it.
01:38
So look, battery is still connected, hence potential difference remains the same.
01:43
Okay? that is basically already given to us in the question.
01:47
So our explanation will be battery is still connected.
02:03
Hence potential difference remains the same.
02:27
V -dash...