00:02
In this problem, we are going to calculate the capacitance per unit length, that is c prime, and it is equals to c divided by l.
00:11
In order to calculate the c prime, let's write the relation for c, as c equals to charge divided by a potential difference between the charge densities.
00:21
We can write a relation for v as v equals to minus integration of 1 to r2, edr, where this is the element.
00:36
Electric field.
00:38
This electric field can be written as e equals to lambda divided by 2 pi epsilon not r into i negative of i because it is directed towards negative x -axis plus lambda divided by 2 pi epsilon not r 2 r prime let's call it r prime and negative i because it is also directed towards negative x -axis.
01:29
Now we can generalize these two relations as equals to minus lambda divided by 2 pi epsilon not into 1 divided by r plus 1 divided by r prime and this r prime is equal to b minus r so this is actually r prime.
02:00
Now by inserting this value into the integral which we have written in the equation number 2, let's call it equation number 1.
02:08
We can write equation number 2 as v equals to minus integration of minus lambda by this 2 pi epsilon 0 into 1 divided by r plus 1 divided by b minus r and then we have 4 d r here...