A rod of length `L` and cross-section area `A` lies along the x-axis between `x=0` and `x=L`. The material obeys Ohm's law and its resistivity varies along the rod according to `rho(x) = rho_0 epsilon^(-x//L)`. The end of the rod `x=0` is at a potential `V_0` and it is zero at `x=L`.
(a) Find the total resistance of the rod and the current in the wire.
(b) Find the electric potential in the rod as a function of `x`.

Answer

`R=(rho_(0)L)/A(1-1/e);I=(V_(0)A)/(rho_(0)L)(e/(e-1)); V=V_(0)((e^(-x//L)-e^(-1)))/(1-e^(-1))`

Explanation

(a) For Resistance
`dR=rho(dx)/A`
`R=int_(0)^(L)rho_(0)e^(-x//L)(dx)/A=(rho_(0))/Aint_(0)^(L)e^(-x//L)dx`
`=(rho_(0))/A[-Le^(-x//L)]_(0)^(L)=(rho_(0))/AL[1-e^(1)]`
`R=(rho_(0))/AL(1-1/e)`
For current
`I=(V_(0)-0)/RimpliesI=(V_(0))/RimpliesI=(V_(o)A)/(rho_(0)L)(e/(e-1))`
(b) For potential at any point
`V=V_(0)-IR`
`V=V_(0)-(V_(0)A)/(rho_(0)L)(1/(1-e^(-1)))xxint_(0)^(x)dR`
`V=V_(0)-(V_(0)A)/(rho_(0)L(1-e^(-1))xx(rho_(0))/Aint_(0)^(x)e^(-x//L) dx`
`V=(V_(0)(e^(-x//L)-1))/(1-e^(-1))`