two spherical boys float half some edge in a calm deep sea their radius a, is very much small er than their separation b, calculate the resistance between them If the resistivity of the is rho water
I hope and pray we are talking about buoys and not boys.
Anyway, here is a solution for totally immersed spheres. If the sphere waterlines are halfway down the sphere, then double the resistance. (you have half as much conductance)
http://www.df.unipi.it/~macchi/TEACHING/FISICA2/PROBLEMS/twospheresistance.pdf
To calculate the resistance between the two spherical boys in water, we need to use the formula for electrical resistance:
Resistance (R) = (ρ * length) / area
Where:
- ρ represents the resistivity of water,
- length is the distance between the two spherical boys, and
- area is the cross-sectional area through which the current passes.
In this case, the length is equal to half the separation between the two spherical boys, so we have:
length = b/2
Now, let's calculate the area. Since the spherical boys are small compared to their separation, we can approximate their shape as spheres. The cross-sectional area of a sphere is given by:
area = π * r^2
However, we have two spheres, so the total area will be:
total area = 2 * π * a^2
Now, we can substitute the values and calculate the resistance. The formula becomes:
Resistance (R) = (ρ * (b/2)) / (2 * π * a^2)
Make sure to substitute the values for a, b, and the resistivity of water (ρ), and then simplify the expression to find the resistance between the two spherical boys.