00:01
In this problem, we are going to calculate the energy in the given region, but this energy is capital u.
00:10
Since we can define the energy density as small u equals to d capital u divided by dv.
00:18
From here we can void the relation for this u as u equals to integration of small u dv.
00:27
Let's call it equation number one.
00:29
Now since we can define the energy density as small u equals to 1 divided by 2, epsilon 0 is squared, where this epsilon not is defined as epsilon not equals to 1 divided by 4 pi k, where this is the columns constant.
00:50
In other case, we have the electric field as e equals to 2k lambda divided by r.
01:01
Where this lambda is the linear charge density.
01:05
Similarly, we can write the relation for dv as dv equals to d into piar square, l, where this r is the radius of the diameter of the wire and l is its length.
01:21
So this can be written as a dv equals to 2 pi r, dr, and there will be l...