Solved Problem on Capacitors
ES
FR
PT-BR
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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 is a perfect sphere of radius equal to 6400 km. What is its capacitance?
The Coulomb Constant in a vacuum is \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
Problem data:
- Capacitance of conductor: C=1 F;
- Radius of the Earth: RT=6400 km;
- Coulomb Constant in a vacuum: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).
a) The capacitance, as a function of the charge Q and the potential V, is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{C=\frac{Q}{V}} \tag{I}
\end{gather}
\]
The electric potential of a spherical conductor, with radius R charged with charge Q, is
given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{V=k_e\frac{Q}{R}} \tag{II}
\end{gather}
\]
substituting the equation (II) into equation (I)
\[
\begin{gather}
C=\frac{Q}{k_e\dfrac{Q}{R}}\\[5pt]
C=\frac{\cancel Q}{k_e}\frac{R}{\cancel Q}\\[5pt]
C=\frac{R}{k_e} \tag{III}\\[5pt]
R=Ck_e
\end{gather}
\]
substituting the problem data
\[
\begin{gather}
R=(1\;\mathrm F)\left(9\times 10^9\;\frac{\mathrm{N.m^2}}{\mathrm{C^2}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{R=9\times 10^{9}\;\text{m}}
\end{gather}
\]
Note: this result shows that 1 farad is a large unit, to have a sphere with this
capacitance it should have a radius of 9×106 km = 9000000 km.
b) First, we convert the radius of the Earth given in kilometers (km) to meters (m) used in the International System of Units (SI)
\[
\begin{gather}
R_{\small T}=6400\;\cancel{\mathrm km}\times\frac{1000\;\mathrm m}{1\;\cancel{\mathrm km}}=6400000\;\mathrm m=6.4\times 10^{6}\;\mathrm m
\end{gather}
\]
Using the equation (III) of the previous item
\[
\begin{gather}
C=\frac{R}{k_e}\\[5pt]
C=\frac{6.4\times 10^6\;\mathrm{\cancel m}}{9\times 10^9\;\frac{\mathrm{N.m^{\cancel 2}}}{\mathrm{C^2}}}\\[5pt]
C\approx 0.7\times 10^6\times 10^{-9}\;\frac{\mathrm{C^2}}{\mathrm{N.m}}\\[5pt]
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{C\approx 0,7\times 10^{-3}\;\mathrm F=0,7\;\mathrm{mF}}
\end{gather}
\]
Note: because 1 farad is a large unit, this result shows us why it is common to use
submultiples like milli (m) = 10−3, micro (μ) = 10−6 or pico
(p) = 10−9.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .