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Two metal balls of the same radius `a` are located in a homongenous poorly conducting medium with resistivity `rho`. Find the resistance of the medium between the balls provided that the separation between them is mush greater than the radius of the ball.

Answer» Let us mentally impart the charge `+q` and `-q` to the balls respectively. The electric field strength at the surface of a ball will be determined only by is own charge and the charge can be considred to be uniformly distributed over the surface, because the other ball is at infinite distance. Magnitude of the field strength is given by,
`E = (q)/(4pi epsilon_(0) a^(2))`
So, current density `j = (1)/(rho) (q)/(4pi epsilon_(0) a^(2))` and electric current
`I = int vec(j). vec(dS) = jS = (q)/(rho 4pi epsilon_(0) a^(2)) = (q)/(rho epsilon_(0))`
But potential difference between the balls,
`varphi_(+) - varphi_(-) = 2 (q)/(4pi epsilon_(0) a)`
Hence, the sought resistanace, varphi `R = (varphi_(+) - varphi_(-))/(I) = (2q//4pi epsilon_(0) a)/(q//rho epsilon_(0)) = (rho)/(2pi a)`


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