Solved Problem on Work and Electric Potential
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Two isolated drops of water, whose radii are 0.4 mm and 0.6 mm, are respectively charged with 8×10−8 C and 1.2×10−7 C. Calculate the potential of the drop formed by the union of the two drops.


Problem data:
  • Radius of drop 1::    r1 = 0.4 mm;
  • Charge 1:    q1 = 8×10−8 C;
  • Radius of drop 2:    r2 = 0.6 mm;
  • Charge 2:    q2 = 1.2×10−7 C;
  • Assuming the system is in a vacuum, we have the Coulomb constant: \( k_{0}=9\times 10^{9}\frac{\;\text{N m}^{2}}{\text{C}^{2}} \).
Problem diagram:
Figure 1

Solution

First, we must convert the units of the drops radii given in millimeters (mm) to meters (m), used in the International System of Units (S.I.)
\[ \begin{gather} 1\;\text{mm}=10^{-3}\;\text{m}\\[8pt] r_{1}=0.4\;\text{mm}=4\times 10^{-1}\times 10^{-3}\;\text{m}=4\times 10^{-4}\;\text{m}\\[8pt] r_{2}=0.6\;\text{mm}=6\times 10^{-1}\times 10^{-3}\;\text{m}=6\times 10^{-4}\;\text{m} \end{gather} \]
The potential of a spherical conductor is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {V=k_{0}\frac{Q}{r}} \tag{I} \end{gather} \]
As the electric charge is conserved, the charge of the drop formed by the union of the primitive drops will be the sum of the charges of each one of the drops.
\[ \begin{gather} Q=q_{1}+q_{2} \tag{II} \end{gather} \]
The volume, v, of the new drop will be the sum of the volumes of the initial drops, v1 and v2
\[ \begin{gather} v=v_{1}+v_{2} \tag{III} \end{gather} \]
The volume of a sphere is given by
\[ \bbox[#99CCFF,10px] {v=\frac{4}{3}\pi r^{3}} \]
The volumes of the initial drops and the final drop of union, considered spherical, are given by
\[ \begin{gather} v=\frac{4}{3}\pi R^{3} \tag{IV-a} \end{gather} \]
\[ \begin{gather} v_{1}=\frac{4}{3}\pi r_{1}^{3} \tag{IV-b} \end{gather} \]
\[ \begin{gather} v_{2}=\frac{4}{3}\pi r_{2}^{3} \tag{IV-c} \end{gather} \]
substituting the three expressions (IV-a), (IV-b), and (IV-c) into expression (III)
\[ \begin{gather} \frac{\cancel{4}}{\cancel{3}}\cancel{\pi} R^{3}=\frac{\cancel{4}}{\cancel{3}}\cancel{\pi} r_{1}^{3}+\frac{\cancel{4}}{\cancel{3}}\cancel{\pi} r_{2}^{3}\\ R^{3}=r_{1}^{3}+r_{2}^{3}\\ R=\sqrt[{3\;}]{r_{1}^{3}+r_{2}^{3}\;} \tag{V} \end{gather} \]
substituting expressions (II) and (V) into expression (I)
\[ V=k_{0}\frac{q_{1}+q_{2}}{\sqrt[{3\;}]{r_{1}^{3}+r_{2}^{3}\;}} \]
substituting the numerical values ​​of the problem
\[ \begin{gather} V=9\times 10^{9}\times \frac{8\times 10^{-8}+1.2\times 10^{-7}}{\sqrt[{3\;}]{\left(4\times 10^{-4}\right)^{3}+\left(6\times 10^{-4}\right)^{3}\;}}\\[5pt] V=9\times 10^{9}\times \frac{8\times 10^{-8}+12\times 10^{-8}}{\sqrt[{3\;}]{64\times 10^{-12}+216\times 10^{-12}\;}}\\[5pt] V=9\times 10^{9}\times \frac{20\times 10^{-8}}{\sqrt[{3\;}]{280\times 10^{-12}\;}}\\[5pt] V=9\times 10^{9}\times \frac{20\times 10^{-8}}{6.5\times 10^{-4}}\\[5pt] V=\frac{180\times 10}{6.5\times 10^{-4}}\\[5pt] V=\frac{180\times 10\times 10^{4}}{6.5}\\[5pt] V=28\times 10\times 10^{4} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {V\simeq 2.8\times 10^{6}\;\text{V}} \]
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