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A charged particle creates an electric field of magnitude 300 \(\mathrm{N} / \mathrm{C}\) at a point \(0.800 \mathrm{~m}\) away. What is the difference in the field magnitude between that point and one at \(0.400 \mathrm{~m}\) ?

Short Answer

Expert verified
The difference in the field magnitude is 900 N/C.

Step by step solution

01

Understand the formula for electric field

The electric field (E) created by a point charge (Q) at a distance (r) is given by the formula: \[ E = \frac{kQ}{r^2} \] where \( k \) is Coulomb's constant \( k = 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2} \). We will use this formula to find the field magnitudes at different distances.
02

Calculate the charge using given conditions

First, we know that at a distance of 0.800 m, the electric field \( E_1 \) is 300 N/C. Using the formula, we have: \[ 300 = \frac{kQ}{0.800^2} \] From this equation, we solve for \( kQ \): \[ kQ = 300 \times 0.800^2 \].
03

Compute kQ

Calculate \( kQ \): \[ kQ = 300 \times 0.64 = 192 \]. Therefore, \( kQ = 192 \, \mathrm{N \, m^2/C} \).
04

Calculate the electric field at 0.400 m

Now that we know \( kQ = 192 \), we calculate the electric field \( E_2 \) at a distance of 0.400 m using the formula: \[ E_2 = \frac{192}{0.400^2} \].
05

Solve for E2

Calculate \( E_2 \): \[ E_2 = \frac{192}{0.16} = 1200 \, \mathrm{N/C} \].
06

Find the difference in the electric field magnitude

The difference in the electric field magnitude between the two points is: \[ \Delta E = E_2 - E_1 = 1200 - 300 = 900 \].

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electromagnetism that describes the force between two charged objects. It states that the force \( F \) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This relationship can be expressed mathematically as: \[ F = \frac{k Q_1 Q_2}{r^2} \]Here, \( k \) is Coulomb's constant, which has a value of \( 8.99 \times 10^9 \ \mathrm{N \, m^2/C^2} \). This constant reflects the strength of the electrostatic force in the vacuum. Coulomb's Law helps us understand how charges interact, and it is crucial for calculating the forces and electric fields in a variety of scenarios. In the context of point charges, this law forms the basis for calculating the electric field one charge can exert on another object or space around it. It shows how distance and charge size are key factors in the electric field's strength.
Point Charge
A point charge is a theoretical model of a charged particle. It assumes that the size of the particle is infinitesimally small compared to the distances involved in the problem. This abstraction allows us to simplify calculations and focus on understanding the electric field's behavior. The concept of a point charge is fundamental in physics education because it is the simplest way to study electric fields and forces.
  • Assumed to have a very small size.
  • Possesses a certain amount of electric charge \( Q \).
  • Used primarily in theoretical physics to model simple systems.
In many practical situations, such as in the original exercise, point charges are hypothetical constructs that enable us to predict and calculate the outcomes of more complex scenarios involving electric fields and forces. By modeling charges this way, we can easily apply Coulomb's Law to understand how they might interact across distances.
Electric Field Magnitude
The electric field magnitude denotes the strength of an electric field at a given point. It is a vector quantity, with both magnitude and direction, often expressed in \( \mathrm{N/C} \) (Newtons per Coulomb). The electric field created by a point charge can be determined using the formula: \[ E = \frac{kQ}{r^2} \]Where \( k \) is Coulomb's constant, \( Q \) is the charge creating the field, and \( r \) is the distance from the charge to the point in question.
  • Calculated by dividing the product of the charge and Coulomb's constant by the square of the distance.
  • Indicates how strong the electric field is at different points in space.
The exercise uses two points at different distances from the charge to show how the electric field's strength varies based on location. This vividly illustrates how the proximity to the charge affects the electric field magnitude, a crucial concept when analyzing electrical interactions in physics.
Distance Dependence
Distance dependence is a key factor in determining the strength of the electric field created by a point charge. The formula \( E = \frac{kQ}{r^2} \) reveals that the electric field strength decreases with the square of the distance from the charge. This means the further you move away from a charge, the weaker its influence becomes.
  • The relationship is inversely proportional to the square of the distance.
  • A closer proximity to the charge results in a stronger electric field.
  • This principle is important when considering fields from multiple charges.
In practical terms, this relation is essential for understanding how fields interact and combine in systems with multiple charges. The exercise demonstrates this by calculating the difference in electric field magnitude at two distinct distances from the charge. It illustrates why understanding the distance factor in electric fields is necessary for predicting how a system's field behaves overall.

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Most popular questions from this chapter

Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were \(2.00 \times 10^{4} \mathrm{~N} / \mathrm{C}\) ? (b) What speed would the proton attain if the field accelerated the proton through a distance of \(1.00 \mathrm{~cm} ?\) (c) How much time would be required?

At some instant the velocity components of an electron moving between two charged parallel plates are \(v_{x}=2.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\) and \(v_{y}=5.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Suppose the electric field between the plates is uniform and given by \(\vec{E}=(120 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{j}}\). In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its \(x\) coordinate has changed by \(2.0 \mathrm{~cm}\) ?

A thin nonconducting rod with a uniform distribution of positive charge \(Q\) is bent into a complete circle of radius \(R\) (Fig. 22-22). The central perpendicular axis through the ring is a \(z\) axis, with the origin at the center of the ring. What is the magnitude of the electric field due to the rod at (a) \(z=0\) and (b) \(z=\infty\) ? (c) In terms of \(R\), at what positive value of \(z\) is that magnitude maximum? (d) If \(R=2.00 \mathrm{~cm}\) and \(Q=5.00 \mu \mathrm{C}\), what is the maximum magnitude?

Charge is uniformly distributed around a ring of radius \(R=4.60 \mathrm{~cm}\), and the resulting electric field magnitude \(E\) is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is \(E\) maximum?

A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, \(3.0 \mathrm{~cm}\) away, in a time \(1.5 \times 10^{-8} \mathrm{~s}\). Just as the electron strikes the second plate, what are its (a) momentum magnitude and (b) kinetic energy? (c) What is the magnitude of the electric field \(\vec{E}\) ?

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