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(II) What is the electric field strength at a point in space where a proton experiences an acceleration of 1.8 million "g's"?

Short Answer

Expert verified
The electric field strength is approximately \(1.85 \times 10^8 \text{ N/C}\).

Step by step solution

01

Identify Given Values

We are given that a proton experiences an acceleration of 1.8 million times the acceleration due to gravity, denoted as \(g\). Since \(g = 9.8 \text{ m/s}^2\), this means the acceleration \(a\) is \(1.8 \times 10^6 \times 9.8 \text{ m/s}^2\).
02

Recall the Formula for Electric Force

The electric force \(F\) acting on a proton in an electric field \(E\) is given by \(F = qE\), where \(q\) is the charge of the proton. The charge of a proton is approximately \(q = 1.6 \times 10^{-19} \text{ C}\).
03

Use Newton's Second Law

Newton's Second Law states \(F = ma\), where \(m\) is the mass of the proton and \(a\) is the acceleration. The mass of a proton is approximately \(m = 1.67 \times 10^{-27} \text{ kg}\).
04

Equate the Forces

Since the electric force is equal to the force from acceleration, we can write \(qE = ma\). Substitute the values: \(1.6 \times 10^{-19} E = 1.67 \times 10^{-27} \times 1.8 \times 10^6 \times 9.8\).
05

Solve for Electric Field Strength \(E\)

Rearrange the equation from Step 4 to find \(E\): \[ E = \frac{m \times a}{q} = \frac{1.67 \times 10^{-27} \times 1.8 \times 10^6 \times 9.8}{1.6 \times 10^{-19}} \] Calculate to find \(E\).
06

Calculation

Perform the calculation:\[ E = \frac{1.67 \times 10^{-27} \times 1.8 \times 10^6 \times 9.8}{1.6 \times 10^{-19}} = 1.85 \times 10^8 \text{ N/C}\]Thus, the electric field strength is approximately \(1.85 \times 10^8 \text{ N/C}\).

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Key Concepts

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

Proton Charge
A proton is a subatomic particle with a positive electric charge. This charge is fundamental to understanding electric forces. The magnitude of a proton's charge is approximately \( 1.6 \times 10^{-19} \text{ C} \) (Coulombs). This small value is critical in calculations involving electric fields, as seen in the exercise. The positive charge of protons causes them to be influenced by electric fields, leading to forces acting upon them.
This charge is a constant used frequently in physics, especially when dealing with particles in electric fields. For example, when you need to find the force exerted on a proton by an electric field, the proton's charge is a key component in the equation \( F = qE \), where \( F \) represents force and \( E \) the electric field strength.
Additionally:
  • The concept of charge explains why protons and electrons are attracted to each other, as protons are positively charged and electrons negatively charged.
  • This attractive force is crucial in forming atoms, and by extension, matter as we know it.
Acceleration
Acceleration describes how quickly the velocity of an object changes with time. In physics, it's a vector quantity, meaning it has both a magnitude and a direction.
In this exercise, a proton is subjected to an acceleration of 1.8 million "g's". Here, "g" refers to the acceleration due to gravity at Earth's surface, which is approximately \( 9.8 \text{ m/s}^2 \). Therefore, an acceleration of 1.8 million "g's" translates to:
  • \( 1.8 \times 10^6 \times 9.8 \text{ m/s}^2 \).
Understanding acceleration is important for solving problems involving motion, such as calculating forces that cause changes in motion, following Newton's Second Law.
Moreover, this concept helps us comprehend the immense forces present in environments like particle accelerators, where subatomic particles can reach extreme accelerations. This is invaluable for studies beyond ordinary terrestrial conditions, advancing areas like high-energy physics.
Newton's Second Law
Newton's Second Law is a fundamental principle in physics formulated by Sir Isaac Newton. It states: the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = ma \)).
This law helps us understand how forces influence motion, describing the relationship between an object's mass, the forces acting upon it, and the resulting acceleration.
In the context of the exercise, we use this law to equate the electric force experienced by the proton with the force arising due to its acceleration:
  • We express it as \( qE = ma \) where \( q \) is the charge of the proton, \( E \) is the electric field strength, \( m \) is the mass of the proton, and \( a \) is the given acceleration.
By substituting known values into the equation, we calculated the strength of the electric field.
Using Newton's Second Law allows us to bridge the gap between theoretical physics and practical calculations, clarifying how changes in forces affect motion.

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Most popular questions from this chapter

(I) What is the electric field at a point when the force on a \(1.25 - \mu \mathrm { C }\) charge placed at that point is \(\vec { \mathbf { F } } = ( 3.0 \hat { \mathbf { i } } - 3.9 \hat { \mathbf { j } } ) \times 10 ^ { - 3 } \mathbf { N } ?\)

(III) A thin glass rod is a semicircle of radius \(R\), Fig. \(21-66\). A charge is nonuniformly distributed along the rod with a linear charge density given by \(\lambda=\lambda_{0} \sin \theta,\) where \(\lambda_{0}\) is a positive constant. Point \(\mathbf{P}\) is at the center of the semicircle. (a) Find the electric field \(\overrightarrow{\mathbf{E}}\) (magnitude and direction) at point P. [Hint: Remember \(\sin (-\theta)=-\sin \theta,\) so the two halves of the rod are oppositely charged.] (b) Determine the acceleration \(\quad\) (magnitude \(\quad\) and direction) of an electron placed at point \(\mathrm{P},\) assuming \(\quad R=1.0 \mathrm{~cm}\) and \(\lambda_{0}=1.0 \mu \mathrm{C} / \mathrm{m}\)

Suppose that electrical attraction, rather than gravity, were responsible for holding the Moon in orbit around the Earth. If equal and opposite charges \(Q\) were placed on the Earth and the Moon, what should be the value of \(Q\) to maintain the present orbit? Treat the Earth and Moon as point particles.

Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about \(150 \mathrm{~N} / \mathrm{C}\) at the Earth's surface and points inward toward the Earth's center. What is the magnitude of the electric charge on the Earth? Is it positive or negative? [Hint: The electric field outside a uniformly charged sphere is the same as if all the charge were concentrated at its center.]

(II) A very thin line of charge lies along the \(x\) axis from \(x = - \infty\) to \(x = + \infty .\) Another similar line of charge lies along the \(y\) axis from \(y = - \infty\) to \(y = + \infty .\) Both lines have a uniform charge per length \(\lambda .\) Determine the resulting electric field magnitude and direction (relative to the \(x\) axis) at a point \(( x , y )\) in the first quadrant of the \(x y\) plane.

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