00:01
Hello students, in this question we have given radius of solid wire 3 mm and 2 coulomb of charge per meter length is given as 2 coulombs.
00:22
Now as gauss's law states that in a closed surface so ds that is equal to q that is charge enclosed.
00:34
Let the element length of element wire let's take this edge.
00:47
So now in radius of 3 mm so radius of 3 mm so there is 2 coulomb charge 2h.
01:01
So we can write pi into 3 mm square that is tends to 2 into h by pi into 3 mm square.
01:14
So pi r square will be tends to 2h into pi r square upon pi into 3 mm square.
01:26
So the charge on radius r so this will be equal to 2 upon 3 mm square.
01:39
Now taking gauss surface of radius so gauss surface of radius so this will be less than 3 mm so we can write d 2 pi r h will be charge that is enclosed.
02:03
So we can write d 2 pi r h that is equal to 2 square h upon 3 mm square.
02:12
So from this we can write d is equal to upon pi 3 mm square.
02:21
Now for radius so that is greater than 3 mm so the charge enclosed will be 2 into h...