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Find the potential difference required to accelerate protons from rest to \(10 \%\) of the speed of light. (At this point, relativistic effects start to become significant)

Short Answer

Expert verified
The potential difference required is approximately \( 4.7 \times 10^6 \) V.

Step by step solution

01

Identify the Problem

We need to find the potential difference that accelerates a proton from rest to 10% of the speed of light. We need to use kinetic and potential energy concepts.
02

Understand Basic Formula

The kinetic energy of a proton when accelerated is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the proton and \( v \) is the velocity (10% of the speed of light here).
03

Calculate Speed of Proton

Calculate the speed of the proton which is 10% of the speed of light using the formula \( v = 0.1c \) where \( c = 3 \times 10^8 \) m/s. This gives us \( v = 3 \times 10^7 \) m/s.
04

Relate Kinetic Energy to Potential Difference

The potential difference required (\( V \)) is related to the kinetic energy by \( eV = KE \), where \( e \) is the charge of the proton (\( 1.6 \times 10^{-19} \) C).
05

Compute Kinetic Energy

Substitute the values into the kinetic energy formula: \( KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3 \times 10^7)^2 \), resulting in \( KE = 7.515 \times 10^{-13} \) J.
06

Solve for Potential Difference

Using \( eV = KE \), substitute \( KE \) and \( e \) into \( V = \frac{KE}{e} = \frac{7.515 \times 10^{-13}}{1.6 \times 10^{-19}} \), which results in \( V = 4.7 \times 10^6 \) V.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. It is an essential concept in physics, particularly when we talk about particles like protons. When a particle is at rest, its kinetic energy is zero; however, when it starts moving, it gains kinetic energy. In this context, the kinetic energy for a proton accelerated to 10% of the speed of light is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Here, \( m \) is the mass of the proton and \( v \) is the velocity achieved. When dealing with subatomic particles, it's crucial to have precise values:
  • The mass of a proton \( m \) is approximately \( 1.67 \times 10^{-27} \) kg.
  • Velocity \( v \) is based on the speed of light \( c = 3 \times 10^8 \) m/s, so 10% is \( v = 3 \times 10^7 \) m/s.
Kinetic energy plays a central role in determining how much work is done in moving a proton up to a certain speed and consequently, how much potential difference is required to accomplish this.
Proton Acceleration
Accelerating a proton is a fundamental exercise in understanding high-energy particle physics. In the given exercise, a proton is accelerated from rest to a significant fraction of the speed of light – specifically to 10%. The potential difference, which essentially is a measure of how much energy per charge is required, directly influences this acceleration:
  • Using the equation \( eV = KE \), we relate the work done by an electric field (potential difference) to the kinetic energy gained by the proton.
  • \( V \) here represents the potential difference in volts.
  • \( e \) is the proton's charge, \( 1.6 \times 10^{-19} \) C.
Accelerating protons to relativistic speeds (a sizeable fraction of the speed of light) allows scientists to probe deeper into the structure of matter and explore the fundamental forces of nature. The calculated potential difference \( V \) of \( 4.7 \times 10^6 \) V tells us the high energy scales necessary for such experiments.
Relativistic Effects
Relativistic effects become important when particles like protons reach substantial velocities, nearing a significant fraction of the speed of light. At these high speeds, classical mechanics (Newtonian physics) cannot accurately describe the behavior of such particles.Key considerations include:
  • The mass of the particle appears to increase from the perspective of an outside observer as the particle speed approaches the speed of light.
  • Time dilation and length contraction also become significant, where time appears to slow down and lengths contract along the direction of motion.
  • The formula \( KE = \frac{1}{2}mv^2 \) starts to become less accurate, and relativistic formulas involving Lorentz factors need to be considered.
For this exercise, we note that when protons are accelerated to 10% of the speed of light, relativistic effects start to appear but are not yet predominant in calculations. However, they highlight the importance of considering physics beyond classical ideas when reaching high speeds.

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Most popular questions from this chapter

The plates of a parallel-plate capacitor have constant charges of \(+Q\) and \(-Q\). Do the following quantities increase, decrease, or remain the same as the separation of the plates is increased? (a) The electric field between the plates; (b) the potential difference between the plates; (c) the capacitance; (d) the energy stored in the capacitor.

Computer Keyboards Many computer keyboards operate on the principle of capacitance. As shown in Figure \(20-16\) each key forms a small parallel-plate capacitor whose separation is reduced when the key is depressed. (a) Does depressing a key increase or decrease its capacitance? Explain. (b) Suppose the plates for each key have an area of \(47.5 \mathrm{mm}^{2}\) and an initial separation of \(0.550 \mathrm{mm}\). In addition, let the dielectric have a dielectric constant of \(3.75 .\) If the circuitry of the computer can detect a change in capacitance of \(0.425 \mathrm{pF}\), what is the minimum distance a key must be depressed to be detected?

Find the difference in electric potential, \(\Delta V=V_{B}-V_{N}\), between the points A and B for the following cases: (a) The electric field does 0.052 J of work as you move a \(+5.7-\mu \mathrm{C}\) charge from \(\mathrm{A}\) to B. (b) The electric field does -0.052 J of work as you move a \(-5.7-\mu C\) charge from \(A\) to \(B\) (c) You perform \(0.052 \mathrm{J}\) of work as you slowly move a \(+5.7-\mu C\) charge from \(A\) to \(B\).

The Bohr Atom The hydrogen atom consists of one electron and one proton. In the Bohr model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius \(0.529 \times 10^{-10} \mathrm{m} .\) What is the electric potential due to the proton at the electron's orbit?

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a \(0.250-\mathrm{kg}\) ball with zero net charge was dropped from rest at a height of \(1.00 \mathrm{m}\). The ball landed 0.552 s later. Next, the ball was given a net charge of \(7.75 \mu \mathrm{C}\) and dropped in the same way from the same height. This time the ball fell for 0.680 s before landing. What is the electric potential at a height of \(1.00 \mathrm{m}\) above the ground on this planet, given that the electric potential at ground level is zero? (Air resistance can be ignored.)

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