Solved Problem on Capacitors
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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 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:    RT=6400 km;
• Coulomb constant:    $$k_{0}=9 \times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}}$$.
Solution

a) The capacitance, as a function of the charge Q and the potential V, is given by
$\bbox[#99CCFF,10px] {C=\frac{Q}{V}} \tag{I}$
The electric potential of a spherical conductor, with radius R charged with charge Q, is given by
$\bbox[#99CCFF,10px] {V=k_{0}\;\frac{Q}{R}} \tag{II}$
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 Q, we obtain
C=Q\frac{R}{k_{0}Q}\\ \begin{align}C=\frac{R}{k_{0}} \tag{III}\end{align}\\ R=Ck_{0}
substituting the problem data
$R=1 \times 9 \times 10^{9}$
$\bbox[#FFCCCC,10px] {R=9 \times 10^{9}\;\text{m}}$
Note: this result shows that 1 farad is a very large unit, to have a sphere with this capacity it should have a radius of 9.106 km = 9000000 km.

b) First, we will convert the radius of the Earth given in kilometers (km) to meters (m) used in the International System of Units (S.I.)
$R_{\text{T}}=6400\;\cancel{\text{km}}\;\frac{1000\;\text{m}}{1\;\cancel{\text {km}}}=6400000\;\text{m}=6.4\times10^{6}\;\text{m}$
Using the expression (III) of the previous item, we have
$C=\frac{R}{k_{0}}\\ C=\frac{6.4\times10^{6}}{9\times10^{9}}\\ C=0.7\times10^{6}\times10^{-9}\\ C=0.7\times10^{-3}$
$\bbox[#FFCCCC,10px] {C=0.7\;\text{mF}}$
Note: Because 1 farad is a very 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.