Solved Problem on Work and Electric Potential

A particle with charge q₁ = 1 μC and a mass of 5 g is launched in the radial direction of another particle, with charge q₂ = 6 μC and fixed, the launch speed is 12 m/s from a distance of 0.3 m. Determine at what distance from the fixed particle the launched particle will have a speed equal to zero. Consider the particle is in the vacuum where the Coulomb constant is \( k_{0}=9\times10^{9}\frac{\;\text{N m}^{2}}{\text{C}^{2}} \) and neglect gravitational effects.

Problem data:

Charge 1: q₁ = 1 μC;
Charge 2: q₂ = 6 μC;
Initial speed of charge 1: v_i = 12 m/s;
Final speed of charge 1: v_f = 0;
Initial distance from charge 1: d_i = 0.3 m;
Mass of charge 1: m = 5 g;
Coulomb constant: \( k_{0}=9\times 10^{9}\frac{\;\text{N m}^{2}}{\text{C}^{2}} \).

Problem diagram:

Charge 2 is positive, q₂ > 0, so it produces an electric field with lines pointing radially outward from the charge. Charge 1 is thrown radially and then follows a field line, as the charge is positive, q₁ > 0, and as \( \vec{F}=q\vec{E} \), the electric force on charge 1 has the same direction as the electric field. The charge is thrown in the opposite direction of the field, it will decelerate due to the electric force until it stops.

Figure 1

Solution

First, let's convert the unit of mass given in grams (g) to kilograms (kg) used in the International System of Units (S.I.).

\[ m=5\;\text{g}=5\times 10^{-3}\;\text{kg} \]

Initially, particle 1 at point A is in the potential, V_A, from this point, when moving to point B, it passes to a potential, V_B, in this displacement there is a work given by

\[ \begin{gather} \bbox[#99CCFF,10px] {W_{A}^{B}=q\left(V_{A}-V_{B}\right)} \tag{I} \end{gather} \]

Figure 2

Using the Work-Kinetic Energy Theorem of Classical Mechanics, the work for a body from A to B is given by the change in Kinetic Energy (Figure 2)

\[ \begin{gather} \bbox[#99CCFF,10px] {W_{A}^{B}=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}} \tag{II} \end{gather} \]

Equating the expressions (I) and (II)

\[ \begin{gather} q_{1}\left(V_{A}-V_{B}\right)=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2} \tag{III} \end{gather} \]

The potential produced by a charge at a position is given by

\[ \bbox[#99CCFF,10px] {V=k_{0}\frac{Q}{r}} \]

the potential produced by charge 2 at points A and B will be

\[ \begin{gather} V_{A}=k_{0}\frac{q_{2}}{d_{i}} \tag{IV-a} \end{gather} \]

\[ \begin{gather} V_{B}=k_{0}\frac{q_{2}}{d_{f}} \tag{IV-b} \end{gather} \]

substituting expressions (IV-a) and (IV-b) into expression (III)

\[ q_{1}\;\left(k_{0}\frac{q_{2}}{d_{i}}-k_{0}\frac{q_{2}}{d_{f}}\right)=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2} \]

factoring k₀q₂ on the left-hand side of the equation and with v_f = 0 on the right-hand side

\[ \begin{gather} q_{1}k_{0}q_{2}\;\left(\frac{1}{d_{i}}-\frac{1}{d_{f}}\right)=\frac{m \times 0^{2}}{2}-\frac{mv_{i}^{2}}{2}\\ \frac{1}{d_{i}}-\frac{1}{d_{f}}=-{\frac{mv_{i}^{2}}{2k_{0}q_{1}q_{2}}}\\ \frac{1}{d_{f}}=\frac{1}{d_{i}}+\frac{mv_{i}^{2}}{2k_{0}q_{1}q_{2}} \end{gather} \]

substituting the numeric values given in the problem

\[ \begin{gather} \frac{1}{d_{f}}=\frac{1}{3\times 10^{-1}}+\frac{5\times 10^{-3}\times 12^{2}}{2\times 9\times 10^{9}\times 1\times 10^{-6}\times 6\times 10^{-6}}\\[5pt] \frac{1}{d_{f}}=\frac{10}{3}+\frac{5\times 144\times 10^{-3}}{108\times 10^{-3}}\\[5pt] \frac{1}{d_{f}}=\frac{10}{3}+\frac{720}{108} \end{gather} \]

simplifying the fraction \( \dfrac{720}{108} \) dividing the denominator and numerator by 36, \( \dfrac{720:36}{108:36}=\dfrac{20}{3} \)

\[ \begin{gather} \frac{1}{d_{f}}=\frac{10}{3}+\frac{20}{3}\\ \frac{1}{d_{f}}=\frac{30}{3}\\ d_{f}=\frac{3}{30} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {d_{f}=0.1\;\text{m}} \]

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