/*! 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 37 Determine the magnitude of the e... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 37

Determine the magnitude of the electric field in vacuum at a distance of \(1.00 \mathrm{~mm}\) from a proton. [Hint: Use \(k_{0}\).]

Short Answer

Expert verified
The magnitude of the electric field is approximately \(1.44 \times 10^3 \text{ N/C}\).

Step by step solution

01

Identify Known Values

Given a proton and a distance of \(1.00 \text{ mm}\), we need to determine the electric field in a vacuum. The charge of a proton is \(+1.602 \times 10^{-19} \text{ C}\), and the distance given is \(1.00 \times 10^{-3} \text{ m}\).
02

Recall the Formula for Electric Field

The electric field \(E\) due to a point charge is given by: \[ E = \frac{k_0 \cdot |q|}{r^2} \]where \(k_0 = 8.988 \times 10^9 \text{ N m}^2/\text{C}^2\), \(q\) is the charge, and \(r\) is the distance from the charge.
03

Substitute Known Values into Formula

Using the formula \(E = \frac{k_0 \cdot |q|}{r^2}\), we substitute:- \(k_0 = 8.988 \times 10^9 \text{ N m}^2/\text{C}^2\)- \(q = 1.602 \times 10^{-19} \text{ C}\)- \(r = 1.00 \times 10^{-3} \text{ m}\)This gives us: \[ E = \frac{8.988 \times 10^9 \cdot 1.602 \times 10^{-19}}{(1.00 \times 10^{-3})^2} \]
04

Calculate Electric Field Magnitude

Perform the calculations:- Calculate \( (1.00 \times 10^{-3})^2 = 1.00 \times 10^{-6} \)- Numerator: \( 8.988 \times 10^9 \cdot 1.602 \times 10^{-19} = 1.44033676 \times 10^{-9} \)- Divide: \( E = \frac{1.44033676 \times 10^{-9}}{1.00 \times 10^{-6}} = 1.44033676 \times 10^{3} \)The electric field magnitude is approximately \(1.44 \times 10^3 \text{ N/C}\).

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Key Concepts

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

Electric Field Formula
The electric field formula is a cornerstone in understanding how electric forces work around charged particles. Consider a point charge, which is a source of electric field. The formula to calculate the electric field \(E\) due to a point charge \(q\) is given by: \[E = \frac{k_0 \cdot |q|}{r^2}\]where:
  • \(k_0\), known as Coulomb's constant, is approximately \(8.988 \times 10^9 \text{ N m}^2/\text{C}^2\).
  • The absolute value \(|q|\) ensures positivity, as electric field magnitude is always non-negative.
  • \(r\) is the distance from the charge, squared.
This formula is essential for calculating the strength of an electric field generated by a single charged particle, like a proton or electron, at a certain distance.
Point Charge
A point charge represents a charged body with its charge concentrated at a single point in space. It is an idealized concept often used in physics to simplify the study of electrostatics. When calculating the electric field, a proton, which has a positive charge, can be treated as a point charge.
  • Protons have a charge of \(+1.602 \times 10^{-19} \text{ C}\).
  • A point charge creates an electric field that radiates outward (or inward for a negative charge) in all directions.
The concept of point charge is used for its simplicity, as it allows easy computation of electric fields using the electric field formula mentioned earlier. It's important to note that the effect of this electric field diminishes with distance.
Coulomb's Constant
Coulomb's constant \(k_0\) is a key component of the electric field formula, acting as a proportionality factor. Its value is approximately \(8.988 \times 10^9 \text{ N m}^2/\text{C}^2\). This constant is derived from Coulomb's Law, which describes the force between two charges.Here is why Coulomb's constant is crucial:
  • It allows us to relate the tiny units of charge (Coulombs) to the much larger magnitudes of force (Newtons).
  • It provides the necessary scale to transform the charge interactions into the electric field strength.
By using \(k_0\), the calculated electric field can be expressed in Newtons per Coulomb (N/C), a standard SI unit for measuring electric field strength.
Distance from Charge
The distance \(r\) from a point charge is a crucial factor when determining the strength of the electric field. In the electric field formula, this distance is squared, which implies an inverse square relationship.Let's break it down:
  • A smaller distance \(r\) results in a larger electric field \(E\), demonstrating a strong interaction.
  • Conversely, as the distance increases, the electric field decreases rapidly, indicating a weaker influence.
  • For instance, if you double the distance from the charge, the electric field strength is reduced by a factor of four.
This inverse square law is a fundamental principle in physics, showcasing how interactions such as gravitational and electric forces diminish quickly with increased separation between objects.

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Most popular questions from this chapter

A charge of \(+6.0 \mu\) C experiences a force of \(2.0 \mathrm{mN}\) in the \(+x\) direction at a certain point in space. (a) What was the electric field at that point before the charge was placed there? ( \(b\) ) Describe the force a \(-2.0 \mu \mathrm{C}\) charge would experience if it were used instead of the \(+6.0 \mu \mathrm{C}\) charge.

Two +400-nC point charges are in vacuum separated by \(20.0 \mathrm{~cm}\). Determine the electric field at a point midway between the charges.

Two point charges, one \(+400.0 \mathrm{nC}\) and the other \(-400.0 \mathrm{nC}\), located \(20.00 \mathrm{~cm}\) to the right of the first, are in vacuum. Determine the electric field (magnitude and direction) at a point midway between the charges.

What is the electrostatic force acting on each of two tiny uniformly charged spheres in vacuum if they both carry \(1.00\) C of charge and they are separated, center to center, by \(1.00 \mathrm{~m}\) ?

Four equal-magnitude \((4.0 \mu \mathrm{C})\) charges in vacuum are placed at the four corners of a square that is \(20 \mathrm{~cm}\) on each side. Find the electric field at the center of the square \((a)\) if the charges are all positive, (b) if the charges alternate in sign around the perimeter of the square, (c) if the charges have the following sequence around the square: plus, plus, minus, minus.
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