A metal ball of radius `a` is located at a distance `l` from an indinite ideally conducting plane. The space around the ball is filled with a homongeous poorly conducting medium with resistivity `rho`. In the case of `a lt lt l` find :
(a) the current density at the conducting plane as a function of distance `r` from the ball if the potentail difference between the ball and the plane is equal to `V` ,
(b) the electric resistance of the medium between the balll and the plane.

Explanation

(a) The potential int eh unshaded region beyond the conductor as the potentail of the given charge and its image and has the form
`varphi = A((1)/(r_(1)) - (1)/(r_(2)))`
where `r_(1), r_(2)` are the distances of the point from the charge and its image. The potential has been taken to be zero on the conducting plane and on the ball
`varphi = A ((1)/(a) - (1)/(2l)) = V`
So `A = Va` In this calculation the conditions `a lt lt l` is used to ignore the variation of `varphi` over the ball.
The electric field at `P` can be calculated similarly. The charge on the ball is
`Q = 4pi epsilon_(0)Va`
and `E_(P) = (Va)/(r^(2)) 2 cos theta = (2a//V)/(r^(3))`
Then `j = (1)/(rho) E = (2alV)/(rho r^(3))` normal to the plane.
(b) The total current flowing into the conducting plane is
`I = int_(0)^(oo) 2pi x dx j = int_(0)^(oo) 2pi x dx j = (2al V)/(rho(pi^(2) + l^(2))^(3//2))`
(On putting `y = x^(2) + l^(2)`)
`I = (2pi alV)/(rho) int_(l^(2))^(oo) (dy)/(y^(3//2)) = (4pi a V)/(rho)`
Hence `R = (V)/(I) = (rho)/(2pi a)`