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: 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
\[ \begin{align} C=\cancel{Q}\frac{R}{k_{0}\cancel{Q}}\\ C=\frac{R}{k_{0}} \quad\;\, \tag{III}\\ R=Ck_{0} \quad\; \end{align} \]
substituting the problem data
\[ R=1 \times 9 \times 10^{9} \]
\[ \bbox[#FFCCCC,10px] {R=9 \times 10^{9}\;\text{m}} \]
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
\[ \begin{gather} C=\frac{R}{k_{0}}\\ C=\frac{6.4\times10^{6}}{9\times10^{9}}\\ C=0.7\times10^{6}\times10^{-9}\\ C=0.7\times10^{-3} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {C=0.7\;\text{mF}} \]
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.


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