/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A uniform electric field exists ... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(1.60 \mathrm{~cm}\) distant from the first, in a time interval of \(3.20 \times 10^{-6} \mathrm{~s} .\) (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Short Answer

Expert verified
The magnitude of the electric field is \(6.56 \times 10^4 \, \mathrm{N/C}\) and the speed of the proton when it hits the negatively charged plate is \(200 \, \mathrm{m/s}\).

Step by step solution

01

Identify the given values

From the exercise, we can identify the given values. The distance between the plates \(d = 1.60 \, \mathrm{cm} = 1.60 \times 10^{-2} \, \mathrm{m}\), the time it takes for the proton to travel \(t = 3.20 \times 10^{-6} \, \mathrm{s}\), and the charge and the mass of the proton \(q = 1.60 \times 10^{-19} \, \mathrm{C}\) and \(m = 1.67 \times 10^{-27} \, \mathrm{kg}\).
02

Find the acceleration of the proton

We know from kinematics that the distance travelled is given by \(d = \frac{1}{2} a t^2\), where a is the acceleration. We can solve the equation for the acceleration: \(a = \frac{2d}{t^2}\). Substituting the given values, we find \(a = \frac{2 \cdot 1.60 \times 10^{-2} \, \mathrm{m}}{(3.20 \times 10^{-6} \, \mathrm{s})^2} = 6.25 \times 10^{7} \, \mathrm{m/s^2}\).
03

Calculate the magnitude of the electric field

The electric field \(E\) is related to the force on the proton and its charge by \(E = \frac{F}{q}\), where \(F\) is the force. Since the force is related to the mass and acceleration of the proton by \(F = ma\), we can substitute this into the previous equation to find \(E = \frac{ma}{q} = \frac{1.67 \times 10^{-27} \, \mathrm{kg} \cdot 6.25 \times 10^{7} \, \mathrm{m/s^2}}{1.60 \times 10^{-19} \, \mathrm{C}} = 6.56 \times 10^4 \, \mathrm{N/C}\).
04

Calculate the speed of the proton

The speed \(v\) of the proton can be found using the equation \(v = at\), where \(a\) is the acceleration and \(t\) is the time. Substituting the given values, we find \(v = 6.25 \times 10^{7} \, \mathrm{m/s^2} \cdot 3.20 \times 10^{-6} \, \mathrm{s} = 200 \, \mathrm{m/s}\).

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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. In this exercise, the proton is set into motion under the influence of an electric field. When a proton is placed in an electric field, it experiences a force along the direction of the field. This is due to the interaction between the proton's positive charge and the electric field's force lines.
The motion is straightforward: the proton starts from rest and accelerates towards the opposite charge, in this case, towards the negatively charged plate. This type of motion is predictable and is a great example of how charged particles interact with electric fields to yield observable behavior.
Newton's Laws
Newton's laws of motion are fundamental to understanding how forces affect motion. Here, Newton's second law is particularly relevant. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, given by the formula:
  • \( F = ma \)
For the proton, we know that the force acting on it is due to the electric field. Therefore, we express this force in terms of the electric field (\( E \)) and the proton's charge (\( q \)):
  • \( F = qE \)
Combining these facts gives us a way to relate the electric field to the proton's acceleration through the equation \( E = \frac{ma}{q} \). This relationship is pivotal in calculating how quickly the proton will speed up in response to the electric field.
Kinematics
Kinematics is the study of motion without considering the forces that cause the motion. To find how far and how fast the proton travels between the plates, we employ kinematics equations.

The formula for the distance \( d \) that an object covers when starting from rest under constant acceleration is:
  • \( d = \frac{1}{2} a t^2 \)
This equation allows us to compute the acceleration \( a \) when we have values for \( d \) and the time \( t \). Furthermore, once the acceleration is determined, the final speed \( v \) of the proton can be calculated using:
  • \( v = at \)
Both these equations illustrate how motion parameters are connected and how kinematics plays a critical role in understanding particle motion under uniform acceleration.
Uniform Electric Field
A uniform electric field is one where the electric force is consistent throughout. This means that a charged particle, such as a proton, will experience the same force magnitude and direction regardless of where it is in the field.

In the scenario provided, the uniform electric field exists between two plane parallel plates, with one positively charged and the other negatively charged. When released from rest at the positively charged plate, the proton moves towards the negatively charged plate under the influence of this electric field. The field provides constant acceleration, creating a steady increase in the proton's speed as it traverses the space between the plates.

The magnitude of the uniform electric field can be established by using the relationship between force, charge, and electric field (\( E = \frac{F}{q} \)). This enables us to quantify the field's influence on the proton, showcasing how a uniform electric field can precisely control the motion of charged particles.

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Most popular questions from this chapter

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m},\) and charge \(q_{2}=+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)

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 \(\alpha=30.0^{\circ}\)

A very long, straight wire has charge per unit length \(3.20 \times 10^{-10} \mathrm{C} / \mathrm{m} .\) At what distance from the wire is the electric- field magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)

Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

Consider an infinite flat sheet with positive charge density \(\sigma\) in which a circular hole of radius \(R\) has been cut out. The sheet lies in the \(x y\) -plane with the origin at the center of the hole. The sheet is parallel to the ground, so that the positive \(z\) -axis describes the "upward" direction. If a particle of mass \(m\) and negative charge \(-q\) sits at rest at the center of the hole and is released, the particle, constrained to the \(z\) -axis, begins to fall. As it drops farther beneath the sheet, the upward electric force increases. For a sufficiently low value of \(m,\) the upward electrical attraction eventually exceeds the particle's weight and the particle will slow, come to a stop, and then rise back to its original position. This sequence of events will repeat indefinitely. (a) What is the electric field at a depth \(\Delta\) beneath the origin along the negative \(z\) -axis? (b) What is the maximum mass \(m_{\max }\) that would prevent the particle from falling indefinitely? (c) If \(m

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