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Ch 23: The Electric Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 23: The Electric FieldProblem 53
Chapter 23, Problem 53

The two parallel plates in FIGURE P23.53 are 2.0 cm apart and the electric field strength between them is 1.0×104 N/C. An electron is launched at a 45° angle from the positive plate. What is the maximum initial speed v0 the electron can have without hitting the negative plate?
Diagram showing two parallel plates 2.0 cm apart with an electron launched at a 45° angle from the positive plate.

Verified step by step guidance
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Step 1: Understand the problem setup. The electron is launched at a 45° angle from the positive plate in a uniform electric field. The goal is to find the maximum initial speed v₀ such that the electron does not hit the negative plate, which is 2.0 cm (0.02 m) away. The electric field strength is given as 1.0×10⁴ N/C.
Step 2: Identify the forces acting on the electron. The electric field exerts a force on the electron given by F = qE, where q is the charge of the electron (-1.6×10⁻¹⁹ C) and E is the electric field strength. This force causes a constant acceleration a = F/m, where m is the mass of the electron (9.11×10⁻³¹ kg).
Step 3: Break the motion into components. The electron's motion can be analyzed in two dimensions: horizontal (x-axis) and vertical (y-axis). The initial velocity v₀ has components v₀ₓ = v₀ cos(45°) and v₀ᵧ = v₀ sin(45°). The vertical motion is influenced by the electric field, while the horizontal motion is unaffected by the field.
Step 4: Use kinematic equations for vertical motion. The vertical displacement y of the electron is given by y = v₀ᵧ t + (1/2) a t², where t is the time of flight and a is the acceleration due to the electric field. To ensure the electron does not hit the negative plate, the maximum vertical displacement y must be less than or equal to 0.02 m.
Step 5: Relate time of flight to horizontal motion. The horizontal displacement x is given by x = v₀ₓ t. Since the electron travels horizontally until it reaches the edge of the plates, the time t can be expressed in terms of the horizontal distance and velocity. Combine this with the vertical motion equation to solve for the maximum initial speed v₀ that satisfies the condition y ≤ 0.02 m.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

An electric field is a region around charged particles where other charged particles experience a force. The strength of the electric field (E) is defined as the force (F) per unit charge (q), measured in newtons per coulomb (N/C). In this scenario, the electric field strength of 1.0×10^4 N/C indicates how strongly the electron will be accelerated towards the negative plate.
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Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to gravitational and other forces. In this case, the electron's trajectory can be analyzed as a projectile launched at a 45° angle, which maximizes its range. Understanding the components of its initial velocity and the effects of the electric field is crucial for determining how far it can travel before hitting the negative plate.
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Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. It involves equations that relate displacement, velocity, acceleration, and time. For the electron, kinematic equations will be used to calculate the maximum initial speed required to ensure it does not collide with the negative plate while moving under the influence of the electric field.
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Related Practice
Textbook Question

CALC A uniform electric field’s strength is increasing with time as E=(1.5×104+(5.0×1010s−1)t)N/CE = (1.5 \(\times\) 10^4 + (5.0 \(\times\) 10^{10}\,\(\text{s}\)^{-1})t)\,\(\text{N/C}\). A proton is released in the field from rest at t = 0. What is the proton’s speed 1.0 μs later?

Textbook Question

Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in FIGURE P23.48. Find an expression for the electric field Ē at the center of the semicircle. Hint: A small piece of arc length Δs spans a small angle Δθ=Δs/R , where R is the radius.

Textbook Question

A problem of practical interest is to make a beam of electrons turn a 90° corner. This can be done with the parallel-plate capacitor shown in FIGURE P23.55. An electron with kinetic energy 3.0×10−17 J enters through a small hole in the bottom plate of the capacitor. Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right? Explain.

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Textbook Question

INT In a classical model of the hydrogen atom, the electron orbits the proton in a circular orbit of radius 0.053 nm. What is the orbital frequency in rev/s? The proton is so much more massive than the electron that you can assume the proton is at rest.

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Textbook Question

An infinite plane of charge with surface charge density 3.2 μC/m2 has a 20-cm-diameter circular hole cut out of it. What is the electric field strength directly over the center of the hole at a distance of 12 cm? Hint: Can you create this charge distribution as a superposition of charge distributions for which you know the electric field?

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Textbook Question

An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and negative charges in opposite directions. The dipole moment of an induced dipole is directly proportional to the electric field. That is, p→=αE→\(\overrightarrow{p}\)=\(\alpha\]\overrightarrow{E}\), where α is called the polarizability of the molecule. A bigger field stretches the molecule farther and causes a larger dipole moment. An ion with charge q is distance r from a molecule with polarizability α. Find an expression for the force E→\(\overrightarrow{E}\)ion on dipole.

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