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Answer :
The final velocity of the charged particle when it is 2.0 m away from the stationary charge is approximately [tex]\(0.574 \ \text{m/s}\).[/tex]
To determine the final velocity of the 1.5 mc charged particle, we can use the principle of conservation of energy. The initial potential energy [tex](\(U_i\))[/tex] is converted into kinetic energy [tex](\(K_f\))[/tex] and the final potential energy [tex](\(U_f\))[/tex]as the particle moves from an initial distance[tex](\(r_i\))[/tex] to a final distance [tex](\(r_f\)).[/tex]
The potential energy at any distance [tex]\(r\)[/tex] between two point charges can be calculated using the formula:
[tex]\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \][/tex]
where [tex]\(k\)[/tex] is Coulomb's constant[tex](\(8.99 \times 10^9 \ \text{N m}^2/\text{C}^2\)), \(q_1\) and \(q_2\)[/tex] are the charges, and [tex]\(r\)[/tex] is the separation between them.
The total energy [tex]\(E\)[/tex] at any point is the sum of kinetic and potential energy:
[tex]\[ E = K + U \][/tex]
At the initial position, all energy is potential [tex](\(E_i = U_i\))[/tex], and at the final position, the energy is the sum of kinetic and potential [tex](\(E_f = K_f + U_f\)).[/tex]
Since the particle starts from rest, [tex]\(K_i = 0\), and \(E_i = U_i\)[/tex]. At the final position, [tex]\(E_f\)[/tex] is the kinetic energy, and [tex]\(U_f\)[/tex] is the potential energy.
Equating the initial and final energies, we get:
[tex]\[ U_i = K_f + U_f \][/tex]
Solving for the final kinetic energy [tex](\(K_f\)):[/tex]
[tex]\[ K_f = U_i - U_f \][/tex]
Using the kinetic energy formula [tex]\(K_f = \frac{1}{2} m v_f^2\), where \(m\)[/tex] is the mass of the particle and [tex]\(v_f\)[/tex] is the final velocity, we can find [tex]\(v_f\).[/tex]
First, let's calculate the initial potential energy [tex]\(U_i\)[/tex] of the system:
[tex]\[ U_i = \frac{k \cdot q_1 \cdot q_2}{r_i} \][/tex]
[tex]\[ U_i = \frac{(8.99 \times 10^9 \ \text{N m}^2/\text{C}^2) \cdot (1.5 \times 10^{-6} \ \text{C}) \cdot (4.4 \times 10^{-6} \ \text{C})}{1.2 \ \text{m}} \][/tex]
[tex]\[ U_i \approx 2.985 \ \text{J} \][/tex]
[tex]\[ U_f = \frac{k \cdot q_1 \cdot q_2}{r_f} \][/tex]
[tex]\[ U_f = \frac{(8.99 \times 10^9 \ \text{N m}^2/\text{C}^2) \cdot (1.5 \times 10^{-6} \ \text{C}) \cdot (4.4 \times 10^{-6} \ \text{C})}{2.0 \ \text{m}} \][/tex]
[tex]\[ U_f \approx 2.247 \ \text{J} \][/tex]
Now, apply the conservation of energy principle:
[tex]\[ K_f = U_i - U_f \][/tex]
[tex]\[ K_f = 2.985 \ \text{J} - 2.247 \ \text{J} \][/tex]
[tex]\[ K_f \approx 0.738 \ \text{J} \][/tex]
Finally, use the kinetic energy formula to find the final velocity [tex](\(v_f\)):[/tex]
[tex]\[ K_f = \frac{1}{2} m v_f^2 \][/tex]
[tex]\[ 0.738 \ \text{J} = \frac{1}{2} (4.3 \ \text{kg}) v_f^2 \][/tex]
Solving for [tex]\(v_f\):[/tex]
[tex]\[ v_f^2 \approx \frac{2 \times 0.738 \ \text{J}}{4.3 \ \text{kg}} \][/tex]
[tex]\[ v_f \approx 0.574 \ \text{m/s} \][/tex]
Therefore, the final velocity of the charged particle when it is 2.0 m away from the stationary charge is approximately [tex]\(0.574 \ \text{m/s}\).[/tex]
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