advertisement

Solved Problem on Capacitors

a) What must be the radius of a conducting sphere, located in the vacuum, so that its capacity is 1 F;

b) Assuming the planet Earth as a perfect sphere of radius equal to 6400 km, what is its capacitance?

Assume the Coulomb constant as. \( k_{0}=9 \times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}} \)

Problem data:

- Capacitance of conductor:
*C*=1 F; - Radius of the Earth:
*R*_{T}=6400 km; - Coulomb constant: \( k_{0}=9 \times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}} \).

a) The capacitance, as a function of the charge

\[
\bbox[#99CCFF,10px]
{C=\frac{Q}{V}} \tag{I}
\]

The electric potential of a spherical conductor, with radius
\[
\bbox[#99CCFF,10px]
{V=k_{0}\;\frac{Q}{R}} \tag{II}
\]

Substituting the expression (II) into (I), we have
\[
C=\frac{Q}{k_{0}\frac{Q}{R}}
\]

inverting the denominator and simplifying the value of the charge
\[
\begin{align}
C=\cancel{Q}\frac{R}{k_{0}\cancel{Q}}\\
C=\frac{R}{k_{0}} \quad\;\, \tag{III}\\
R=Ck_{0} \quad\;
\end{align}
\]

substituting the problem data
\[
R=1 \times 9 \times 10^{9}
\]

\[
\bbox[#FFCCCC,10px]
{R=9 \times 10^{9}\;\text{m}}
\]

b) First, we will convert the radius of the Earth given in kilometers (km) to meters (m) used in the

\[
R_{\text{T}}=6400\;\cancel{\text{km}}\;\frac{1000\;\text{m}}{1\;\cancel{\text {km}}}=6400000\;\text{m}=6.4\times10^{6}\;\text{m}
\]

Using the expression (III) of the previous item, we have
\[
\begin{gather}
C=\frac{R}{k_{0}}\\
C=\frac{6.4\times10^{6}}{9\times10^{9}}\\
C=0.7\times10^{6}\times10^{-9}\\
C=0.7\times10^{-3}
\end{gather}
\]

\[
\bbox[#FFCCCC,10px]
{C=0.7\;\text{mF}}
\]

advertisement

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .