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.
inner conductor
C
outer conductor
C

Answer 1

Concepts and reason

The concept of cylindrical capacitance is required to solve the problem.

Initially, the value of capacitance per unit length is calculated by using the inner and outer radius of the cylindrical conductor and the length of the cylindrical capacitor. Then, the charge on the inner and outer conductor is determined by using the relation between capacitance, voltage between the conductors and charge.

Fundamentals

The expression for capacitance per unit length is,

$\frac{C}{L} = \frac{{2\pi k{\varepsilon _0}}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$

Here, C is the capacitance, L is the length of the capacitor, k is the dielectric constant, ${\varepsilon _0}$ is the permittivity of free space, ${r_1}$ is the inner radius, and ${r_2}$ is the outer radius.

The magnitude of charge on a capacitor is determined by using the formula,

$Q = CV$

Here, C is the capacitance and V is the voltage between the inner and outer conductor.

(a)

The capacitance per unit length is calculated by using the formula,

$\frac{C}{L} = \frac{{2\pi k{\varepsilon _0}}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$

Substitute 1 for k, $8.85 \times {10^{ - 12}}{\rm{ F/m}}$ for ${\varepsilon _0}$, 1.5 mm for ${r_1}$, and 3.4 mm for ${r_2}$ in the above equation $\frac{C}{L} = \frac{{2\pi k{\varepsilon _0}}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$.

$\begin{array}{c}\\\frac{C}{L} = \frac{{2\left( {3.14} \right)\left( 1 \right)\left( {8.85 \times {{10}^{ - 12}}{\rm{ F/m}}} \right)}}{{\ln \left( {\frac{{3.4{\rm{ mm}}}}{{1.5{\rm{ mm}}}}} \right)}}\\\\ = 6.8 \times {10^{ - 11}}{\rm{ F/m}}\\\end{array}$

(b)

The magnitude of charge on each conductor is determined by using the formula,

$Q = CV$

Substitute $\frac{{2\pi k{\varepsilon _0}L}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$ for C in equation $Q = CV$.

$Q = \frac{{2\pi k{\varepsilon _0}LV}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$

Here, L is the length of the capacitor, k is the dielectric constant, ${\varepsilon _0}$ is the permittivity of free space, ${r_1}$ is the inner radius, and ${r_2}$ is the outer radius.

Substitute 1 for k, $8.85 \times {10^{ - 12}}{\rm{ F/m}}$ for ${\varepsilon _0}$, 1.5 mm for ${r_1}$, 3.4 mm for ${r_2}$, 350 mV for V in equation $Q = \frac{{2\pi k{\varepsilon _0}LV}}{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}$.

$\begin{array}{c}\\Q = \frac{{2\left( {3.14} \right)\left( 1 \right)\left( {8.85 \times {{10}^{ - 12}}{\rm{ F/m}}} \right)\left( {3.0{\rm{ m}}} \right)\left( {350{\rm{ mV}}\left( {\frac{{{{10}^{ - 3}}{\rm{ V}}}}{{1{\rm{ mV}}}}} \right)} \right)}}{{\ln \left( {\frac{{3.4{\rm{ mm}}}}{{1.5{\rm{ mm}}}}} \right)}}\\\\ = 7.1 \times {10^{ - 11}}{\rm{ C}}\\\end{array}$

Charge on the inner conductor is,

${Q_{{\rm{inner}}}} = + 7.1 \times {10^{ - 11}}\;{\rm{C}}$

Charge on the outer conductor is,

${Q_{{\rm{outer}}}} = - 7.1 \times {10^{ - 11}}{\rm{ C}}$

Ans: Part a

The value of capacitance per unit length is $6.8 \times {10^{ - 11}}{\rm{ F/m}}$.