00:01
The radius of the rod r is equal to 4 .80 cm or 0 .048 meter.
00:13
Charge per unit length is given.
00:15
So, lambda is equal to 28 .2 nanoculum per meter or 28 .2 into 28 .2 into 10 to the power minus 9 coulum per meter or 28 .2 into 10 to the power minus 9 coulum per meter.
00:31
To find the electric field at a distance d from the rod, we consider a cylindrical surface, so that is a gaussian surface.
00:46
So it is an imaginary surface.
00:54
So the length of the gaussian surface is l and the distance from the rod is d.
01:08
So this is the small patch area.
01:15
Now, from the goss law, we can write closed integral of electric field dot the area vector is equal to total charge n closed by the area divided by epsilon not.
01:30
Now, q is the lambda into l, so that is charge n closed by the given cylinder.
01:41
Now, e.
01:45
Dot d s is equal to lambda l divided by epsilon not.
01:50
And eds cos 0 is equal to lambda l divided by epsilon 0.
02:04
So eds is equal to lambda l divided by epsilon not.
02:10
Here integration of ds is area of the curved surface.
02:15
So that is given by e into 2 pi into d into l is equal to lambda l divided by abjel.
02:25
By isolating e, that is electric field, we get 1 divided by 4 pi epsilon 0 into 2 lambda divided by d...