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A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. Ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s},\) and \(\alpha=30.0^{\circ}\)

Short Answer

Expert verified
The proton will descend a maximum distance of \(h_{max}\) below its initial position and will return to its initial height after covering a horizontal distance of \(d\). The specific numerical values of \(h_{max}\) and \(d\) will depend on the values given for the electric field \(E\), initial velocity \(v_{0}\), and initial angle \(\alpha\).

Step by step solution

01

Understand the basics

A proton moving in an electric field behaves similarly to a projectile under gravity, but in the opposite direction because the electric field is directed upwards. Due to this field, the proton will have an acceleration that equals \(E/|e|\), where \(|e|\) is the magnitude of the charge on a proton.
02

Calculate the maximum height

The maximum vertical distance \(h_{max}\) is given by \(h_{max} = v_{0}^2 * sin^2(\alpha) / (2a)\), where the acceleration \(a = E/|e|\), and \(\alpha\) (alpha) is the initial angle to the horizontal. The proton will reach the maximum height when it momentarily comes to a stop before reversing direction, as a direct consequence of the force applied by the electric field.
03

Calculate the horizontal distance

The time the proton takes to return to its original height is calculated using the equation \(t = 2* v_{0}* sin(\alpha)/a\). Because the velocity is constant in the horizontal direction, the horizontal distance \(d\) the proton travels before returning to its original height can be given by \(d = v_{0}* cos(\alpha) * t\)
04

Finding numerical values

To find the numerical values for \(h_{max}\) and \(d\), substitute the given values \(E = 500 N/C\), \(|e|=1.602 * 10^{-19} C\), \(v_{0} = 4.00 * 10^{5} m/s\), and \(\alpha = 30^\circ\) into the equations previously derived for \(h_{max}\) and \(d\). In this step, remember to convert \(\alpha\) into radians.

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Key Concepts

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

Electric Field
An electric field is a region of space where charged particles experience a force. This force can change the motion of the particles, much like gravity does.

Electric fields are measured in newtons per coulomb (N/C), indicating the force per unit charge. In our exercise, the electric field points vertically upwards with a magnitude of 500 N/C. When a charged particle, like a proton, enters an electric field, it experiences a force due to its charge.

The force exerted on the proton can be determined by multiplying the field strength \(E\) by the proton's charge \(|e|\). This results in acceleration in the same direction as the force. In this case, the acceleration \(a = E/|e|\) where \(|e|\) = 1.602 × 10^{-19} C, the charge of a proton. This relationship underpins the motion of the proton in the electric field we examine in problems like these.
Projectile Motion
Projectile motion refers to the path an object follows as it moves through space under the influence of gravity or other forces such as an electric field.

In projectile motion, an object is projected into the air at an angle with an initial velocity. During its flight, it experiences uniform acceleration, which in usual circumstances is gravity. However, in this exercise, gravity is ignored and replaced by the acceleration caused by the electric field.

Key characteristics of projectile motion include:
  • The horizontal motion is uniform, meaning it has a constant velocity.
  • The vertical motion is uniformly accelerated, meaning it changes velocity at a constant rate.
  • The trajectory of projectile motion typically forms a parabolic shape.
By analyzing these components, we can predict the path and determine distances, like the horizontal distance \(d\) and maximum height \(h_{max}\) reached by the projectile.
Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion.

It focuses on various parameters like displacement, velocity, acceleration, and time. In the context of a moving proton in an electric field, these parameters are crucial in determining the object's behavior as it traverses its path.

We split the proton's motion into two components:
  • Horizontal: The velocity remains constant since the electric field only affects the vertical motion, making horizontal calculations straightforward.
  • Vertical: The acceleration caused by the electric field changes the vertical velocity over time.This motion can be described by the kinematic equations, such as calculating the time to reach maximum height using \(t = 2 \cdot \frac{v_{0} \cdot \sin(\alpha)}{a}\).
Understanding these components helps predict and calculate various aspects of motion, providing a complete picture of the proton's journey.
Physics Problem Solving
Solving physics problems like the one in this exercise involves understanding the underlying principles and applying them with accuracy.

Key steps include:
  • Identifying the given variables: Such as initial velocity \(v_{0}\), angle \(\alpha\), and field strength \(E\).
  • Understanding relevant physics principles: Recognize that the electric field acts like gravity here, exerting a force on the proton.
  • Applying equations: Utilize kinematic equations and understand the unique role the electric field plays in altering motion characteristics like acceleration.
  • Calculating results: Substituting given numerical values into the derived formulas helps determine specific measurements like maximum descent and horizontal displacement.
Breaking down problems step by step and using structured methods ensures a clear pathway from understanding to solution, crucial for physics problem-solving success.

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Most popular questions from this chapter

A thin disk with a circular hole at its center, called an \(a n-\) nulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. \(\mathbf{P 2 1 . 8 7}\) ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus. (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

Two small spheres, each carrying a net positive charge, are separated by \(0.400 \mathrm{~m}\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge \(q_{1}\) ) at the origin and the other sphere (charge \(q_{2}\) ) at \(x=+0.400 \mathrm{~m}\). Available to you are a third sphere with net charge \(q_{3}=4.00 \times 10^{-6} \mathrm{C}\) and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\) -axis at \(x=0.200 \mathrm{~m} ;\) you measure the net force on it to be \(4.50 \mathrm{~N}\) in the \(+x\) -direction. Then you move the third sphere to \(x=+0.600 \mathrm{~m}\) and measure the net force on it now to be \(3.50 \mathrm{~N}\) in the \(+x\) -direction. (a) Calculate \(q_{1}\) and \(q_{2}\). (b) What is the net force (magnitude and direction) on \(q_{3}\) if it is placed on the \(x\) -axis at \(x=-0.200 \mathrm{~m} ?\) (c) At what value of \(x\) (other than \(x=\pm \infty\) ) could \(q_{3}\) be placed so that the net force on it is zero?

A charge \(+Q\) is located at the origin, and a charge \(+4 Q\) is at distance \(d\) away on the \(x\) -axis. Where should a third charge, \(q\), be placed, and what should be its sign and magnitude, so that all three charges will be in equilibrium?

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of \(2.50 \mathrm{~mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

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