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An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu s\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Short Answer

Expert verified
(a) Field magnitude is \(5.69 \times 10^2 \text{ N/C}\), directed downward. (b) Gravity is negligible compared to the electric force.

Step by step solution

01

Given Data

The initial velocity of the electron is 0 because it is released from rest. The displacement of the electron is given as \( s = 4.50 \text{ m} \) and the time taken is \( t = 3.00 \times 10^{-6} \text{ s} \). We know the charge and mass of an electron are \( e = 1.60 \times 10^{-19} \text{ C} \) and \( m = 9.11 \times 10^{-31} \text{ kg} \).
02

Calculate Acceleration

Using the equation of motion: \( s = ut + \frac{1}{2} a t^2 \), where \( u = 0 \), we have:\[4.50 = 0 + \frac{1}{2} a (3.00 \times 10^{-6})^2\]Solving for \( a \):\[a = \frac{2 \times 4.50}{(3.00 \times 10^{-6})^2} = 1.00 \times 10^{12} \text{ m/s}^2\]
03

Determine Electric Field Magnitude

Using the formula \( F = ma \) and \( F = eE \), equate them: \( eE = ma \). Solving for \( E \):\[E = \frac{ma}{e} = \frac{9.11 \times 10^{-31} \times 1.00 \times 10^{12}}{1.60 \times 10^{-19}} = 5.69 \times 10^2 \text{ N/C}\]
04

Direction of Electric Field

Since the electron is negatively charged and moves upward, the electric field must be directed downward.
05

Justifying Gravity Neglect

Calculate gravitational force: \( F_g = mg \), where \( g = 9.81 \text{ m/s}^2 \):\[F_g = 9.11 \times 10^{-31} \times 9.81 = 8.93 \times 10^{-30} \text{ N}\]Compare it to the electric force: \( F_e = eE = 1.60 \times 10^{-19} \times 5.69 \times 10^2 = 9.10 \times 10^{-17} \text{ N} \).Since \( F_e \gg F_g \), we can ignore the effects of gravity.

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Key Concepts

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

Electron Motion
Imagine an electron, a tiny particle with a negative charge, moving against the invisible currents of an electric field. When it starts from a complete stop, it begins gaining speed due to the force acting upon it. This scenario mirrors what happens when you release an electron in a uniform electrical field. The electron initially at rest will accelerate in one direction, allowing you to track its movement over a certain distance and time.
This motion showcases how electrons react to surrounding electric forces, a vital concept in understanding how electric fields influence charged particles. When an electron is set in motion within an electric field, it gains kinetic energy, causing it to accelerate. The trajectory and velocity can be calculated using various physics principles, prominently the equations of motion.
Studying electron motion provides insights into phenomena like electrical currents and circuits, where electrons flow in specified directions. This movement underlines the foundational behavior of charged particles, which is crucial for fields like electronics, electromagnetism, and quantum physics.
Equations of Motion
Equations of motion are mathematical tools that describe the movement of objects. They precisely link various aspects of motion, such as displacement, velocity, acceleration, and time. In the case of an electron within an electric field, these equations help us calculate how far and how fast an electron travels after being accelerated by an electric force.
For instance, the standard equation of motion used here, \( s = ut + \frac{1}{2} a t^2 \), relates the displacement of an object \( s \), its initial velocity \( u \), its acceleration \( a \), and the time \( t \) over which it has been moving. Given that our electron is released from rest, its initial velocity is zero, which simplifies the equation. By rearranging this formula, we can solve for any unknown quantity if the others are provided.
These equations form the backbone of classical mechanics, allowing us to understand and predict the behavior of moving bodies, not just in electric fields but in a wide range of contexts from everyday physics to advanced engineering.
Electric Force vs Gravitational Force
Electric and gravitational forces are two fundamental interactions that influence object motion, but they differ significantly in strength and behavior.
- **Electric Force:** This arises from interactions between charged particles. Its strength depends on the charge of the particles and the electric field they are in. In our exercise, the electric force much outweighs the gravitational pull. This makes it the dominant factor affecting electron motion. - **Gravitational Force:** It is the attraction between two masses, often noticeable at larger scales. For tiny particles like electrons, gravity is practically negligible compared to electric forces. In fact, calculations show that the electric force acting on the electron is many magnitudes stronger than gravity. When comparing the two types of forces, it becomes evident why in scenarios involving subatomic particles, the role of gravity is often overlooked. The immense strength of the electric forces simply dwarfs the gravitational forces, making the latter irrelevant for practical computations.
Ignoring Gravity in Electrostatics
In electrostatics, the forces being studied are primarily those arising from electric charges. When dealing with atomic-scale particles like electrons, the effect of gravity is so minor that it's typically ignored.
This simplification stems from the fact that gravitational forces become negligible in comparison to electric forces at such small scales. For example, in our electron motion exercise, the electric force calculated is orders of magnitude larger than the gravitational force acting on the electron. Thus, gravity's impact is so insignificant that it doesn’t alter the results. In practice: - **Focus on Electric:** Calculations concentrate on electric interactions, ensuring precision and reducing unnecessary complexity. - **Simplifies Analysis:** Ignoring gravity streamlines problem-solving processes in electrostatics, providing clear insights into how electric fields behave without gravitational disturbances. By disregarding gravity, scientists and engineers can streamline calculations, focusing solely on the more impactful electric forces which dictate the behavior of charged particles in these fields.

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Most popular questions from this chapter

A small sphere with mass \(9.00 \mu \mathrm{g}\) and charge \(-4.30 \mu \mathrm{C}\) is moving in a circular orbit around a stationary sphere that has charge \(+7.50 \mu \mathrm{C}\). If the speed of the small sphere is \(5.90 \times 10^{3} \mathrm{~m} / \mathrm{s}\), what is the radius of its orbit? Treat the spheres as point charges and ignore gravity.

An infinite sheet with positive charge per unit area \(\sigma\) lies in the \(x y\)-plane. A second infinite sheet with negative charge per unit area \(-\sigma\) lies in the \(y z\)-plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors \(\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{J}}\), and \(\hat{\boldsymbol{k}}\).

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive \(x\)-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$ F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right] $$ (c) Show that if \(a \gg L, \quad\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} . \quad\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots\), valid for \(|z| \ll 1\). Carry all expansions to at least order \(L^{2} / a^{2}\).) Interpret this result.

(a) What must the charge (sign and magnitude) of a \(1.3 \mathrm{~g}\) particle be for it to remain stationary when placed in a downward-directed electric field of magnitude \(650 \mathrm{~N} / \mathrm{C}\) ? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

A particle has charge \(-3.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C}\) ?

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