Solved Problem on Capacitors

a) What must be the radius of a conducting sphere in the vacuum so that its capacity is 1 F?
b) Assuming the planet Earth is a perfect sphere with a 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: R_E=6400 km;
Coulomb Constant in a vacuum: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \).

Solution:

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

\[ \begin{gather} R=(1\;\mathrm F)\left(9\times 10^9\;\mathrm{\small{\frac{N.m^2}{C^2}}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {R=9\times 10^{9}\;\mathrm m} \end{gather} \]

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)

\[ \begin{gather} R_{\small E}=6400\;\mathrm{\cancel{km}}\times\frac{1000\;\mathrm m}{1\;\mathrm{\cancel{km}}}=6400000\;\mathrm m=6.4\times 10^{6}\;\mathrm m \end{gather} \]

Using the equation (III) of the previous item

\[ \begin{gather} C=\frac{R}{k_e} \\[5pt] C=\frac{6.4\times 10^6\;\mathrm{\cancel m}}{9\times 10^9\;\mathrm{\frac{N.m^{\cancel 2}}{C^2}}} \\[5pt] C\approx 0.7\times 10^6\times 10^{-9}\;\mathrm{\small{\frac{C^2}{N.m}}} \\[5pt] \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {C\approx 0.7\times 10^{-3}\;\mathrm F=0.7\;\mathrm{mF}} \end{gather} \]

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.

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