Question

A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and the entirecapacitor is 3.0 m long.(a) What is the capacitance per unit length?F/m(b) The potential of the inner conductor is 350 mV higher than thatof the outer conductor. Find the charge (magnitude and sign) onboth conductors.

Answer 1

Answer:

Part(a): The capacitance per unit length of the capacitor is $\bf{1.56 \times 10^{-10}}~F~m^{-1}$.

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

$C = \dfrac{2 \pi \epsilon_{0}}{log(b/a)}$

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

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

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

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

$Q = C~V~L$

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

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

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