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Chapter 21: Problem 27

A proton is traveling horizontally to the right at \(4.50 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Short Answer

Expert verified
For proton, the electric field is \(3.16 \times 10^4\ \mathrm{N/C}\) to the left. Time to stop is \(1.42 \times 10^{-9} \mathrm{s}\). For electron, field is the same magnitude, opposite direction.

Step by step solution

01

Analyze the Problem

We need to stop a proton over a distance of 3.20 cm using an electric field. Given the proton's initial velocity of \(4.50 \times 10^6 \ \mathrm{m/s}\), we need to find the electric field that will decelerate it to rest.
02

Understand the Forces Involved

The force on the proton by the electric field \(F\) is given by \(F = qE\), where \(q\) is the charge of the proton \(1.60 \times 10^{-19}\ \mathrm{C}\), and \(E\) is the electric field strength.
03

Calculate the Required Deceleration

Using the kinematic equation \(v_f^2 = v_i^2 + 2a x\), with \(v_f = 0\), \(v_i = 4.50 \times 10^6\ \mathrm{m/s}\), and \(x = 3.20 \ \mathrm{cm} = 0.032\ \mathrm{m}\), solve for \(a\). This gives us \(a = \frac{-(v_i^2)}{2x}\).
04

Substitute and Solve for Electric Field

Using \(F = ma\) and \(F = qE\), substitute \(ma = qE\) to get \(E = \frac{ma}{q}\). Substitute the known values to find \(E\).
05

Determine Direction of Electric Field

Since the proton is traveling to the right and needs to stop, the electric field must point to the left to decelerate the proton with a force opposite to its motion.
06

Calculate Time to Stop

Use the equation \(v_f = v_i + at\). Solving for \(t\) gives \(t = \frac{-v_i}{a}\), using \(a\) calculated previously.
07

Analyze the Scenario for an Electron

Repeat the calculation for an electron using its charge \(-1.60 \times 10^{-19}\ \mathrm{C}\). Determine that a field of the same magnitude as calculated for the proton is needed, but it should point in the opposite direction (to the right).
08

Calculate Electric Field for Electron

Since the magnitudes are the same as for the proton, the calculations remain unchanged except for considering the direction opposite to that for the proton.

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Key Concepts

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

Proton Motion
Protons are positively charged particles found in the nucleus of an atom. When a proton moves through an electric field, its motion can be affected significantly.
  • Because protons have a positive charge, they experience force in the direction of the electric field. However, if the field is in the opposite direction of the motion, the proton will slow down.
  • In the context of our exercise, we need an electric field to bring a moving proton to rest. This requires the field to exert a force opposite to the proton's initial direction of motion.
  • The direction of this electric force will be crucial in determining how quickly the proton can stop.
By understanding proton motion in an electric field, we can manipulate conditions to either speed up or slow down the proton based on our needs.
Kinematics in Physics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It's crucial in calculating various parameters like velocity, acceleration, and displacement.To bring a proton to rest using an electric field, we utilize kinematic equations.
  • The equation we use is: \( v_f^2 = v_i^2 + 2a x \)
  • Here, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(x\) is the displacement. In our exercise, \(v_f\) becomes zero since the proton comes to a halt.
  • Rearranging the equation helps us solve for acceleration, \(a = \frac{-v_i^2}{2x}\).
This acceleration is due to the force exerted by the electric field, and understanding this helps us relate the motion with the forces at play.
Electric Force
Electric force is the interaction between two charged objects. This force is what we use to manipulate the motion of charged particles like protons and electrons.
  • For a proton, the electric force is calculated using the equation \( F = qE \). Here, \( q \) is the charge, and \( E \) is the electric field.
  • In our scenario, this force is used to stop the proton by exerting it in the opposite direction of motion.
  • We combine this with the mass \(m\) of the proton to determine the electric field required: \(E = \frac{ma}{q}\).
This concept is pivotal to controlling particle motion, such as stopping a proton over a given distance. The direction and magnitude of the electric field determine the effectiveness of this control.

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Most popular questions from this chapter

Two positive point charges \(q\) are placed on the \(x\) -axis, one at \(x=a\) and one at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0\) . (b) Derive an expression for the electric field at points on the \(x\) -axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x\) , for values of \(x\) between \(-4 a\) and \(+4 a\) .

Two small aluminum spheres, each having mass 0.0250 \(\mathrm{kg}\) are separated by \(80.0 \mathrm{cm} .\) (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 \(\mathrm{g} / \mathrm{mol}\) , and its atomic number is \(13 . )\) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{N}\) (roughly 1 ton \() ?\) Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Two particles having charges \(q_{1}=0.500 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC}\) are separated by a distance of 1.20 \(\mathrm{m}\) . At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

Two identical spheres are each attached to silk threads of length \(L=0.500\) in and hung from a common point (Fig. 21.44\()\) . Each sphere has mass \(m=8.00 \mathrm{g}\) . The radius of each sphere isvery small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1}\) , and the other a different positive charge \(q_{2}\) ; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw afree-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\)Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

(a) Sketch the electric field lines for an infinite line of charge. You may find it helpful to show the field lines in a plane containing the line of charge in one sketch and the field lines in a a plane perpendicular to the line of charge in a second sketch. (b) Explain how your sketches show (i) that the magnitude \(E\) of the electric field depends only on the distance \(r\) from the line of charge and (ii) that \(E\) decreases like \(1 / r .\)
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