00:01
Two long coaxial cylindrical metal tubes with an inner radius a and an outer radius b stand vertically in a tank of dielectric oil.
00:10
The inner one is maintained at a potential v and the outer one is grounded.
00:15
To what height does the oil rise in the space between the tubes? so we need to find the height, which means we can equate two forces.
00:28
First, for the net upward force, the force is one half v squared dc over dh and the gravitational force downwards is mg which is row pi b squared minus a squared g h okay so using this to help us first we need to figure out what c is for our force appear so so for the part in air, the field is 2 lambda over 4 pi epsilon s, which means the potential is 2 lambda over 4 pi epsilon knot and the natural log of b over a.
01:37
When we're talking about the oil, d is 2 lambda prime over 4 pi s, which makes the field 2 lambda prime over 4 pi epsilon s.
01:59
Notice not epsilon not.
02:02
And this makes v2 lambda over 4 pi epsilon and then ln of b over a.
02:13
Looking at these two and comparing them, we see that lambda over epsilon not is equal to, who i dropped my prime, lambda prime over epsilon...