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Two protons, starting several meters apart, are aimed directly at each other with speeds of \(2.00 \times 10^5\) m\(/\)s, measured relative to the earth. Find the maximum electric force that these protons will exert on each other.

Short Answer

Expert verified
The maximum electric force between the protons is calculated using Coulomb's law at the distance of closest approach.

Step by step solution

01

Define the Variables

The speed of each proton is given as \(v = 2.00 \times 10^5\ \text{m/s}\) and this is the initial speed relative to earth. Two protons have equal mass \(m_p = 1.67 \times 10^{-27}\ kg\) and charge \(e = 1.60 \times 10^{-19}\ C\). We want to find the maximum electric force that they exert on each other.
02

Apply Conservation of Energy

The initial kinetic energy \(K_i\) of the system comes from both protons. Therefore, \(K_i = 2 \times \frac{1}{2}mv^2 = mv^2\). This energy will be converted into electrical potential energy (\(U\)) at the point of closest approach.
03

Calculate Initial Kinetic Energy

The initial kinetic energy for one proton is \(K = \frac{1}{2} mv^2\). Adding up for both protons gives us \(K_i = mv^2 = (1.67 \times 10^{-27}\ kg)(2.00 \times 10^5\ \text{m/s})^2\). Calculate this to find \(K_i\).
04

Set Kinetic Energy Equal to Potential Energy

The point at which the two protons stop moving toward each other and start to move away is when their kinetic energy is fully converted into electric potential energy, which is given by \(U = \frac{k e^2}{r}\), where \(k = 8.99 \times 10^9\ \text{N m}^2/\text{C}^2\) is Coulomb's constant, \(e\) is the charge of a proton, and \(r\) is the separation distance.
05

Solve for Distance of Closest Approach

Set the calculated initial kinetic energy equal to the potential energy expression: \[ mv^2 = \frac{k e^2}{r} \]Solve for \(r\).
06

Calculate Maximum Electric Force

The force between the protons at the distance of closest approach can be calculated using Coulomb's law:\[ F = \frac{k e^2}{r^2} \]Substitute the value of \(r\) obtained from the previous step into the equation to find the maximum force.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The conservation of energy is a foundational concept in physics that states that energy can neither be created nor destroyed, only transformed from one form to another. In the exercise involving two protons, this principle is crucial in understanding how their energy transitions between kinetic and electric potential energy.
  • Initially, each proton has a certain amount of kinetic energy due to its motion.
  • As they approach each other, this kinetic energy is gradually converted into electric potential energy.
  • The point at which the protons cease their approach marks the total conversion of kinetic energy into potential energy.
Conservation of energy ensures that the sum of kinetic and electric potential energy remains constant throughout the protons' motion. This is fundamental to solving the exercise, allowing us to set these energies equal at the closest point to find the distance of closest approach.
Coulomb's Law
Coulomb's Law is a key equation used to describe the electric force between two charged objects. It is given by:\[ F = \frac{k e^2}{r^2} \]where:
  • \(F\) is the force between the charges,
  • \(k\) is Coulomb's constant \(8.99 \times 10^9 \text{N m}^2/\text{C}^2\),
  • \(e\) is the charge of a proton \(1.60 \times 10^{-19}\),
  • \(r\) is the distance between the charges,
Coulomb's Law tells us that the electric force is inversely proportional to the square of the distance between the charges and exponentially greater as they are closer. This is significant in the exercise, as the maximum electric force occurs at the closest distance between the protons. By determining this minimum distance from the conservation of energy, we can calculate the maximum force exerted.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion, and is calculated with the formula:\[ K = \frac{1}{2} mv^2 \]where:
  • \(K\) is the kinetic energy,
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object,
In the given problem, each proton starts with a significant kinetic energy due to their high speed of \(2.00 \times 10^5\ \text{m/s}\). The total initial kinetic energy contributes to the electric potential energy at the point of closest approach.
This conversion illustrates how energy cannot just disappear. Instead, it transforms into a different form that influences the behavior of the system, like repelling forces that stop the two protons as they approach each other.
Electric Potential Energy
Electric potential energy is the energy stored between two charged objects due to their positions in an electric field. It is expressed as:\[ U = \frac{k e^2}{r} \]where:
  • \(U\) is the electric potential energy,
  • \(k\) is Coulomb's constant,
  • \(e\) is the charge of a proton,
  • \(r\) is the separation distance between the charges.
In the context of this exercise, as the protons move closer, their initial kinetic energy converts into electric potential energy until motion stops entirely. This is when the conversion is complete and the electric potential energy is at its maximum, describing the point of closest approach.Understanding electric potential energy helps visualize at which point two charged particles will repel each other enough to stop moving closer, an integral aspect of solving the problem at hand.

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Most popular questions from this chapter

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10}\) m (a typical atomic distance) to \(3.00 \times 10^{-15}\) m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

Charge \(Q =\) 5.00 mC is distributed uniformly over the volume of an insulating sphere that has radius \(R =\) 12.0 cm. A small sphere with charge \(q = +\)3.00 \(\mu\)C and mass 6.00 \(\times 10^{-5}\) kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

An object with charge \(q = -6.00 \times 10^{-9}\) C is placed in a region of uniform electric field and is released from rest at point \(A\). After the charge has moved to point \(B\), 0.500 m to the right, it has kinetic energy \(3.00 \times 10^{-7}\) J. (a) If the electric potential at point \(A\) is \(+\)30.0 V, what is the electric potential at point \(B\)? (b) What are the magnitude and direction of the electric field?

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