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Chapter 23: Problem 66

A proton with a speed of \(1.23 \cdot 10^{4} \mathrm{~m} / \mathrm{s}\) is moving from infinity directly toward a second proton. Assuming that the second proton is fixed in place, find the position where the moving proton stops momentarily before turning around.

Short Answer

Expert verified
Answer: Approximately \(5.94 * 10^{-15}\) meters.

Step by step solution

01

Identify the initial and final conditions

Initially, the moving proton has a certain kinetic energy due to its speed, and potential energy due to its distance from the second proton (assumed to be an infinite distance). Finally, the moving proton will stop momentarily, meaning its speed (and thus kinetic energy) will be zero, but it will be at a distance r from the second proton, where the potential energy between the two is at a maximum.
02

Apply conservation of energy

To apply the conservation of energy, we equate the initial total energy to the final total energy. The initial total energy consists of kinetic energy and potential energy, while the final energy consists only of potential energy since the proton stops momentarily (kinetic energy is zero). Initial energy = Final energy \(K_{initial} + U_{initial} = U_{final}\)
03

Choose the appropriate formulas for kinetic and potential energy

Since the moving proton has an initial kinetic energy due to its speed, K_initial can be calculated using the formula: \(K_{initial} = \frac{1}{2} mv^2\) Where m is the mass of the proton and v is its initial speed. For potential energy, we need to choose a formula that involves distance between two charged particles. Since the particles are protons, they both have a positive charge (e). The formula for potential energy is: \(U = k*\frac{q_1 * q_2}{r}\) Where q1 and q2 are the charges of the protons, r is the distance between them, and k is Coulomb's constant.
04

Write the equation for conservation of energy with the chosen formulas

Insert the formulas for the initial and final energies into the conservation of energy equation: \(\frac{1}{2} mv^2 + 0 = k*\frac{e^2}{r}\)
05

Solve for r (the distance of the momentarily stopped proton)

Now, we have the equation needed to find the distance r. Plug in the values for the moving proton's speed, mass, and charge: \(\frac{1}{2} * (1.67 * 10^{-27} kg) * (1.23 * 10^4 m/s)^2 = (8.99 * 10^9 N m^2/C^2) * \frac{(1.6 * 10^{-19} C)^2}{r}\) Solve for r: \(r \approx 5.94 * 10^{-15} m\) So, the moving proton will stop momentarily at a distance of approximately \(5.94 * 10^{-15}\) meters from the fixed proton before turning around.

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Most popular questions from this chapter

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.

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