A particle (charge = Q) is kept in a fixed position at point P, and a second particle (charge = q) is released from rest when it is a distance R from P. If Q = +2.0 mC, q = −1.5 mC, and R = 30 cm, what is the kinetic energy of the moving particle after it has moved a distance of 10 cm?
A) 60 kJ
B) 45 kJ
C) 75 kJ
D) 90 kJ
E) 230 kJ

Question

Quizplus · Accepted Answer

The kinetic energy gained by the moving particle can be calculated using the conservation of energy principle. The potential energy at the initial position (distance R from P) is converted into kinetic energy as the particle moves closer. The potential energy difference between the two positions (initial and final) is given by the formula:$$ \Delta U = k \cdot \frac{Qq}{r_1} - k \cdot \frac{Qq}{r_2} $$where $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$ is Coulomb's constant, $Q$ and $q$ are the charges, and $r_1$ and $r_2$ are the initial and final distances from point P, respectively.Given $Q = +2.0 \, \text{mC} = 2.0 \times 10^{-3} \, \text{C}$, $q = -1.5 \, \text{mC} = -1.5 \times 10^{-3} \, \text{C}$, $R = 30 \, \text{cm} = 0.3 \, \text{m}$, and the particle moves 10 cm closer, so the final distance $r_2 = R - 0.1 \, \text{m} = 0.2 \, \text{m}$.$$ \Delta U = 8.99 \times 10^9 \cdot \left( \frac{(2.0 \times 10^{-3}) \cdot (-1.5 \times 10^{-3})}{0.3} - \frac{(2.0 \times 10^{-3}) \cdot (-1.5 \times 10^{-3})}{0.2} \right) $$$$ \Delta U = 8.99 \times 10^9 \cdot \left( -\frac{3 \times 10^{-6}}{0.3} + \frac{3 \times 10^{-6}}{0.2} \right) $$$$ \Delta U = 8.99 \times 10^9 \cdot (-10 \times 10^{-6} + 15 \times 10^{-6}) $$$$ \Delta U = 8.99 \times 10^9 \cdot 5 \times 10^{-6} $$$$ \Delta U = 44.95 \times 10^3 \, \text{J} $$Rounding to the nearest option gives us 45 kJ, which is option B.

A Particle (Charge = Q) Is Kept in a Fixed