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\(\bullet\) The electric field caused by a certain point charge has a mag- nitude of \(6.50 \times 10^{3} \mathrm{N} / \mathrm{C}\) at a distance of 0.100 \(\mathrm{m}\) from the charge. What is the magnitude of the charge?

Short Answer

Expert verified
The magnitude of the charge is approximately \( 7.23 \times 10^{-9} \) 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 \( r \) can be calculated using the formula: \[ E = \frac{k \cdot q}{r^2} \] where \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).
02

Rearrange the formula to solve for charge

To find the charge \( q \), rearrange the formula to solve for \( q \): \[ q = \frac{E \cdot r^2}{k} \]
03

Substitute the known values into the equation

Substitute the given values: \( E = 6.50 \times 10^3 \, \text{N/C} \), \( r = 0.100 \, \text{m} \), and \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \) into the equation: \[ q = \frac{6.50 \times 10^3 \cdot (0.100)^2}{8.99 \times 10^9} \]
04

Calculate the magnitude of the charge

Calculate \( q \): First, compute \( r^2 = (0.100)^2 = 0.010 \). Then compute the charge: \[ q = \frac{6.50 \times 10^3 \times 0.010}{8.99 \times 10^9} = \frac{65}{8.99 \times 10^9} \approx 7.23 \times 10^{-9} \, \text{C} \]
05

Present the final answer

The magnitude of the charge is approximately \( 7.23 \times 10^{-9} \) Coulombs.

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Key Concepts

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

Electric Field
The electric field is a vector quantity that describes the influence a point charge has on surrounding charges. It represents the force per unit charge experienced by a small positive test charge placed in the field. The concept is fundamental in electrostatics, serving as a bridge between the forces acting between charges and the charged objects themselves. The electric field, denoted by \( E \), can be calculated using the formula: \( E = \frac{F}{q_0} \), where \( F \) is the force acting on the test charge \( q_0 \).
An important characteristic of the electric field is its directionality. It points in the direction a positive test charge would move if placed within the field. In practical terms, this means that around positive charges, the electric field radiates outward, while around negative charges, it converges inward. Electric field strength is measured in newtons per coulomb (N/C). This indicates how much force a charge of one coulomb would experience at a specific location in the field.
When dealing with point charges, the electric field decreases in strength as you move further away from the charge. This behavior illustrates the inverse-square nature of the electric field, which we explore further in the concept of Coulomb's Law.
Point Charge
A point charge is an idealized model of a charged object where the size of the charge is much smaller than the distance to the point of interest. We often represent atoms or molecules as point charges in physics, especially when the detailed distribution of charge within an object does not significantly affect the calculations. Point charges are simple to work with mathematically, enabling us to derive fundamental laws like Coulomb's Law.
In the context of an electric field, a point charge generates an electric field that radiates outward or inward depending on the charge's sign. For example:
  • Positive point charge: The electric field radiates outward in all directions.
  • Negative point charge: The electric field points inward toward the charge.
The concept of a point charge is crucial for understanding the behavior of electric fields and electrostatic forces. It simplifies the analysis of complex systems by focusing on the basic interactions between charges.
Coulomb's Law
Coulomb's Law describes the electrostatic force between two point charges. It is a cornerstone of electrostatics, helping us understand how charged particles interact over distances. According to Coulomb's Law, the magnitude of the force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[ F = \ rac{k \, |q_1 \, q_2|}{r^2} \] Here, \( k \) is Coulomb's Constant, which we'll discuss further in the next section.
The force's magnitude is directly proportional to the product of the charges' magnitudes and inversely proportional to the square of the distance between them. This relationship highlights how the force diminishes rapidly as the distance increases, a characteristic shared by gravitational forces as well.
Coulomb's Law also tells us about the direction of the force. It is attractive if the charges are of opposite signs and repulsive if the charges are of the same sign. This law is essential for calculating electric fields, as it helps determine how each charge contributes to the net field at any point in space.
Coulomb's Constant
Coulomb's Constant, often denoted by \( k \), is a proportionality factor in Coulomb's Law. It quantifies the strength of the electrostatic force between two point charges. The value of Coulomb's Constant is approximately \( 8.99 \times 10^9 \, \text{N} \, \cdot \, \text{m}^2 / \text{C}^2 \), a crucial number in the electric field and force calculations.
The presence of Coulomb's Constant in our formulas acts as a scale factor, allowing us to measure electrostatic interactions in the metric system. It combines factors like permittivity of free space to provide a precise standard for these calculations.
When working with real-world problems, knowing the value of \( k \) is essential. It lets us translate the mathematical expression of Coulomb's Law into tangible forces that we can measure in experiments or use in simulations. Without \( k \), our understanding of electrostatics would lack consistency and scale across different scenarios.

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Most popular questions from this chapter

\(\bullet$$\bullet\) In a certain region of space, the electric field \(E\) is uniform; i.e., neither its direction nor its magnitude changes in the region. (a) Use Gauss's law to prove that this region of space must be electrically neutral; that is, there must be no charge in this region. (b) Is the converse true? That is, in a region of space where there is no charge, must \(\vec{E}\) be uniform? Explain.

\(\bullet$$\bullet\) An irregular neutral conductor has a hollow cavity inside of it and is insulated from its surroundings. An excess charge of \(+16 \mathrm{nC}\) is sprayed onto this conductor. (a) Find the charge on the inner and outer surfaces of the conductor. (b) Without touching the conductor, a charge of - 11 \(\mathrm{nC}\) is inserted into the cavity through a small hole in the conductor. Find the charge on the inner and outer surfaces of the conductor in this case.

\(\bullet$$\bullet\) An average human weighs about 650 \(\mathrm{N}\) . If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650 -N weight?

\(\bullet\) . Electric field of axons. A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0\(\mu \mathrm{m}\) in diameter, and meas- urements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e )\) enter during this process. Although the axon is a long cylinder, the charge does not all enter every- where at the same time. A plausible model would be a a series of nearly point charges moving along the axon. Let us look at a 0.10 mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10 \(\mathrm{mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 \(\mathrm{cm}\) below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0\(\mu \mathrm{N} / \mathrm{C}\) . How far from this segment of axon could a shark be and still detect its electric field?

\(\bullet\) As you walk across a synthetic-fiber rug on a cold, dry win- ter day, you pick up an excess charge of \(-55 \mu \mathrm{C}\) . (a) How many excess electrons did you pick up? (b) What is the charge on the rug as a result of your walking across it?

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