Answer:

Part(a): The capacitance per unit length of the capacitor is

$1.56\times {10}^{-10}F{m}^{-1}$

Part(b): The charge in the inner shell is $+1.6\times {10}^{-10}C$ and the charge of the outer shell is $-1.6\times {10}^{-10}C$

Explanation:

Part(a):

The capacitance (C) per unit length of a cylindrical capacitor is given by

$C=\frac{2\pi {\u03f5}_{0}}{\mathrm{log}(b/a)}$

where '${\u03f5}_{0}$' is the permittivity of free space, 'b' is the radius of the outer shell and 'a' is the radius of the inner shell.

Given, b=3.4mm=0.0034 and a=1.5mm=0.0015 mm. We know, ${\u03f5}_{0}=8.854\times {10}^{-12}F{m}^{-1}$. Substituting the values in the above equation,

$C=\frac{2\pi \times 8.85\times {10}^{-12}F{m}^{-1}}{\mathrm{log}(0.0034/0.0015)}=1.56\times {10}^{-10}F{m}^{-1}$

Part(b):

As the voltage on the inner conductor is higher than that of the outer conductor, positive charge resides on the inner shell and negative charge resides on the outer shell.

The charge 'Q' of a capacitor of length 'L' having capacitance 'C' and potential difference 'V' is given by

Q=CVL

Given, $V=350mV=350\times {10}^{-3}V$ and L=3.0m. Substituting these values in the above expression

$Q=1.56\times {10}^{-10}F{m}^{-1}\times 350\times {10}^{-3}V\times 3m=1.6\times {10}^{-10}C$

The charge in the inner shell is $+1.6\times {10}^{-10}C$ and the charge of the outer shell is $-1.6\times {10}^{-10}C.$