Solved Problem on Coulomb's Law

Two spheres are charged with positive charges Q and 3Q, they are placed at a distance d in a vacuum, and between them, we have a force of magnitude F. Then the spheres are placed in contact and separated by a distance 2d. Determine the magnitude of the new repulsive force as a function of F.

Problem data:

Charge of sphere 1: Q;
Charge of sphere 2: 3Q;
Distance between spheres: d;
Magnitude of the force between the spheres: F;
Coulomb constant: k₀.

Solution

Initially, the spheres are placed at a distance d from each other, as both have positive charges, and a repulsive force of magnitude F acts between them (Figure 1).

Figure 1

Coulomb's Law is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{E}=k_{0}\frac{|Q|\;|q|}{r^{2}}} \tag{I} \end{gather} \]

substituting problem data

\[ \begin{gather} F=k_{0}\frac{Q\times 3Q}{d^{2}}\\ F=3k_{0}\frac{Q^{2}}{d^{2}} \tag{II} \end{gather} \]

Then the spheres are brought into contact the electric charges are distributed equally between them

\[ \begin{gather} \frac{Q+3Q}{2}=\frac{4Q}{2}=2Q \end{gather} \]

and then the spheres are separated and placed at a new distance of 2d (Figure 2)

Figure 2

The magnitude of the final force, F_f, between the charges will be found by applying expression (I)

\[ \begin{gather} F_{f}=k_{0}\frac{2Q\times 2Q}{(2d)^{2}}\\ F_{f}=k_{0}\frac{4Q^{2}}{4d^{2}}\\ F_{f}=k_{0}\frac{Q^{2}}{d^{2}} \tag{III} \end{gather} \]

Comparing expressions (II) and (III)

\[ \begin{gather} F=3F_{f} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{f}=\frac{F}{3}} \end{gather} \]

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