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Chapter 23: Problem 69

A particle with a charge of \(+5.0 \mu C\) is released from rest at a point on the \(x\) -axis, where \(x=0.10 \mathrm{~m}\). It begins to move as a result of the presence of a \(+9.0-\mu C\) charge that remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point \(x=0.20 \mathrm{~m} ?\)

Short Answer

Expert verified
Answer: The kinetic energy of the particle at the instant it passes the point x = 0.20 m is 2.022 x 10^{-3} J.

Step by step solution

01

Determine the initial and final positions of the particle

The initial position of the particle is x=0.10 m, and the final position is x=0.20 m.
02

Calculate the initial potential energy

The potential energy between two charged particles is given by the formula: \(U = \frac{kq_1q_2}{r}\), where \(k=8.9875\times 10^9 Nm^2/C^2\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges of the particles, and \(r\) is the distance between them. At the initial position (x=0.10 m), the potential energy is: \(U_i = \frac{(8.9875\times 10^9 Nm^2/C^2)(5.0\times 10^{-6} C)(9.0\times 10^{-6} C)}{0.10 m}\). Calculate the initial potential energy: \(U_i = 4.044\times 10^{-3} J\).
03

Calculate the final potential energy

At the final position (x=0.20 m), the potential energy is: \(U_f = \frac{(8.9875\times 10^9 Nm^2/C^2)(5.0\times 10^{-6} C)(9.0\times 10^{-6} C)}{0.20 m}\). Calculate the final potential energy: \(U_f = 2.022\times 10^{-3} J\).
04

Apply the conservation of mechanical energy

Since the initial kinetic energy of the particle is zero, the initial mechanical energy is equal to the initial potential energy: \(E_{mech_i} = U_i\). The final mechanical energy is the sum of the final potential energy and the final kinetic energy: \(E_{mech_f} = U_f + KE_f\). Applying the conservation of mechanical energy, we get: \(E_{mech_i} = E_{mech_f}\), or \(U_i = U_f + KE_f\).
05

Calculate the final kinetic energy

Solve for the final kinetic energy: \(KE_f = U_i - U_f\). Calculate the final kinetic energy: \(KE_f = 4.044\times 10^{-3} J - 2.022\times 10^{-3} J\). \(KE_f = 2.022\times 10^{-3} J\). The kinetic energy of the particle at the instant it passes the point \(x = 0.20 m\) is \(2.022 \times 10^{-3} J\).

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