The capacitance per unit length is `C/L=Q/(VL)` . So lets find the voltage of this configuration. In order to do that use Gausses' law to find `E` from the charge distribution. Make a Gaussian cylinder.

`int int_A E*dr=Q_(enc)/epsilon_0`

We know that we will enclose part of the charge distribution when `0ltrlta` so make a Gaussian cylinder. Make a surface `r` between `0` and `a`.` `

`E 2pi rL=(rho*pi r^2 L)/epsilon_0`

`E(r)=(rho r )/(2 epsilon_0)`

`0ltrlta.`in the +r direction

Now we must find the field between `altrltb` . Make a cylindrical surface a distance `r` between `a` and `b` .

`int int_A E*dr=Q_(enc)/epsilon_0`

`E (2pi r L)=(rho pi a^2 L)/epsilon_0`

`E(r)=(rho a^2)/(2epsilon_0 r)` in the +r direction for `altrltb`

Now since surface `b` is a grounded conductor it will take up an equal and opposite total charge. Therefore Gausses' law says that the E-field is zero for `rgtb` .

Since we now know `E` for all of space we can integrate for the potential of this system.

`V=Delta V=V(b)-V(a)=-int_a^b E*dr=-int_0^a (rho r )/(2 epsilon_0)dr-int_a^b (rho a^2)/(2epsilon_0 r)dr`

`V=-(rho)/(4 epsilon_0 ) r^2|_0^a -(rho a^2)/(2 epsilon_0)ln(r)|_a^b`

`V=-[(rho)/(4 epsilon_0 )a^2+(rho a^2)/(2 epsilon_0)ln(b/a)]`

`V=[-(rho a^2)/(4epsilon_0)(2ln(b/a)+1)]`

Although, `V(b)ltV(a)` here so for the purposes of this problem make V positive (can only have a positive `C` ).

`V=Delta V=V(a)-V(b)=(rho a^2)/(4epsilon_0)(2ln(b/a)+1)`

`C/L=Q/(VL)=Q/((rho a^2)/(4epsilon_0)(2ln(b/a)+1)L)=Q/((Q a^2)/(pi a^2 L 4 epsilon_0)*(2ln(b/a)+1)L)`

`C/L=1/(4pi epsilon_0)(2ln(b/a)+1)^-1`

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