Solved Problem on Coulomb's Law

Consider two particles A and B, in a vacuum, far from any other body. Particle A is fixed and has charge +Q. Particle B is in Circular Motion with a center at A and a radius of r, has mass m and charge −q. Neglecting gravitational force, determine the speed of B.

Problem data:

Charge of particle A: +Q;
Charge of particle B: −q;
Mass of particle B: m;
Radius of path B: r;
Coulomb constant: k₀.

Problem diagram:

The tangential speed, \( \vec{v} \), and the electric force, \( {\vec{F}}_{E} \), act on particle B.

Figure 1

Solution

Coulomb's Law is given by

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

\[ \begin{gather} F_{E}=k_{0}\frac{|\;Q\;||\;-q\;|}{r^{2}}\\ F_{E}=k_{0}\frac{Qq}{r^{2}} \tag{I} \end{gather} \]

Particle B is in Circular Motion, Newton's Second Law for circular motion is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{cp}=ma_{cp}} \tag{II} \end{gather} \]

the centripetal acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {a_{cp}=\frac{v^{2}}{r}} \tag{III} \end{gather} \]

substituting expression (III) into expression (I)

\[ \begin{gather} F_{cp}=m\frac{v^{2}}{r} \tag{IV} \end{gather} \]

The charges have opposite signs, so the electric force between them is attractive, the only force that acts on the particle at B, substituting expression (I) into expression (IV)

\[ \begin{gather} k_{0}\frac{Qq}{r^{2}}=m\frac{v^{2}}{r}\\ v^{2}=k_{0}\frac{Qq}{r^{\cancel{2}}}\frac{\cancel{r}}{m} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {v=\sqrt{k_{0}\frac{Qq}{mr}}} \]

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