Question

A cylindrical capacitor consists of a solid inner conducting core with radius 0.280 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 39.0 pF.

A) Calculate the outer radius of the hollow tube.
for this i got r=0.332 cm

B) When the capacitor is charged to 120 V, what is the charge per unit length \lambda on the capacitor?

\lambda= _______C/m

Answer 1

Concepts and reason

The question is based upon capacitance of cylindrical capacitor.

For the radius of the capacitor, use the expression of capacitance.

For the linear charge density of the capacitor, first find the charge on the capacitor and use the final expression for the charge density.

Fundamentals

The capacitance C of a cylindrical capacitor is given as follows:

$C = \frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{{{{\log }_e}\left( {\frac{{{R_2}}}{{{R_1}}}} \right)}}$

Here, ${\varepsilon ^\circ }$is the permittivity of free space and ${\varepsilon _r}$is the relative permittivity of the medium between the two cylinders. The symbols ${R_2}$, ${R_1}$are the outer and inner radius respectively and $L$is the length of the cylinder.

The charge Q on the capacitor is,

$Q = CV$

Here, V is the potential difference applied across the capacitor.

The charge per unit length $\lambda$ of the capacitor is,

$\lambda = \frac{Q}{L}$

(A)

Find the outer radius of the capacitor.

The capacitance of cylindrical capacitor is given by following expression:

$C = \frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{{{{\log }_e}\left( {\frac{{{R_2}}}{{{R_1}}}} \right)}}$

Rewrite the expression for the outer radius ${R_2}$as follows:

$\begin{array}{l}\\C = \frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{{{{\log }_e}\left( {\frac{{{R_2}}}{{{R_1}}}} \right)}}\\\\{\log _e}\left( {\frac{{{R_2}}}{{{R_1}}}} \right) = \frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{C}\\\\\left( {\frac{{{R_2}}}{{{R_1}}}} \right) = {e^{\frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{C}}}\\\\{R_2} = {R_1}{e^{\frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{C}}}\\\end{array}$

Substitute $39.0{\rm{ pF}}$ for C, $12.0{\rm{ cm}}$ for L , $8.85 \times {10^{ - 12}}{\rm{ }}{{\rm{N}}^{ - 1}} \cdot {{\rm{C}}^2}/{\rm{ }}{{\rm{m}}^2}$ for $2\pi {\varepsilon ^\circ }$, 1 for ${\varepsilon _r}$ and $0.280{\rm{ cm}}$ for ${R_1}$ in equation ${R_2} = {R_1}{e^{\frac{{2\pi {\varepsilon ^\circ }{\varepsilon _r}L}}{C}}}$.

$\begin{array}{c}\\{R_2} = \left( {0.280{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)\exp \left\{ {\frac{{2\pi \left( {8.85 \times {{10}^{ - 12}}{\rm{ }}{{\rm{N}}^{ - 1}} \cdot {{\rm{C}}^{\rm{2}}}{\rm{/}}{{\rm{m}}^{\rm{2}}}} \right)\left( 1 \right)\left( {12.0{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}{{\left( {39.0{\rm{ pF}}} \right)\left( {\frac{{1 \times {{10}^{ - 12}}{\rm{ F}}}}{{1{\rm{ pF }}}}} \right)}}} \right\}\\\\ = 3.32 \times {10^{ - 2}}{\rm{ m}}\\\\{\rm{ = 0}}{\rm{.332 cm}}\\\end{array}$

(B)

Find the charge per unit length on the capacitor.

The charge per unit length of the capacitor is,

$\lambda = \frac{Q}{L}$

The charge Q on the capacitor is,

$Q = CV$

Here, V is the potential difference applied across the capacitor.

So, the linear charge density or charge per unit length is,

$\lambda = \frac{{CV}}{L}$

Substitute $39.0{\rm{ pF}}$ for C, $12.0{\rm{ cm}}$ for L and ${\rm{120 V}}$ for $V$in equation $\lambda = \frac{{CV}}{L}$.

$\begin{array}{c}\\\lambda = \frac{{\left( {39.0{\rm{ pF}}} \right)\left( {\frac{{{{10}^{ - 12}}{\rm{ F}}}}{{1{\rm{ pF}}}}} \right)\left( {{\rm{120 V}}} \right)}}{{\left( {12.0{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}\\\\ = 3.9 \times {10^{ - 8}}{\rm{ C/m}}\\\end{array}$

Ans: Part A

The outer radius of the hollow tube in the capacitor is ${\rm{0}}{\rm{.332 cm}}$ .