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Chapter 21: Problem 26

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 m directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)

Short Answer

Expert verified
(a) 431.04 N/C, downward; (b) 0.750 m.

Step by step solution

01

Understand the Formula for Electric Field

The magnitude of the electric field \(E\) due to a point charge \(q\) is given by the formula \(E = \frac{k |q|}{r^2}\), where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(q\) is the charge in coulombs, and \(r\) is the distance from the charge in meters.
02

Calculate Electric Field Magnitude for Part (a)

Convert the charge from nanocoulombs to coulombs: \(-3.00 \, \text{nC} = -3.00 \times 10^{-9} \, \text{C}\).Using the formula for electric field, \(E = \frac{(8.99 \times 10^9) \times 3.00 \times 10^{-9}}{(0.250)^2}\).Calculate the magnitude as \(E \approx 431.04 \, \text{N/C}\).
03

Determine the Direction for Part (a)

Since the charge is negative, the electric field direction is toward the charge. Therefore, the electric field at the point 0.250 m above the charge is directed downward.
04

Set Up Equation for Part (b)

We need to find the distance \(r\) such that the electric field \(E = 12.0 \, \text{N/C}\).Use the formula \(12 = \frac{(8.99 \times 10^9) \times 3.00 \times 10^{-9}}{r^2}\).
05

Solve for Distance in Part (b)

Rearrange the equation: \(r^2 = \frac{(8.99 \times 10^9) \times 3.00 \times 10^{-9}}{12}\).Solve for \(r\) to find \(r \approx 0.750 \, \text{m}\).

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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 fundamental in calculating the electric forces between point charges, as well as the electric field created by such charges. It states that the magnitude of the electric force \(F\) between two point charges is directly proportional to the product of the magnitudes of the charges \(q_1\) and \(q_2\), and inversely proportional to the square of the distance \(r\) between them.
The formula is expressed as \[ F = \frac{k |q_1 q_2|}{r^2}, \]
where \(k\) is Coulomb鈥檚 constant, \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).

While Coulomb's Law calculates force, it is also crucial for finding the electric field \(E\) created by a point charge \(q\).
  • To find the electric field, we modify the formula to \(E = \frac{k |q|}{r^2}\).
  • This describes how a point charge influences a region in space.
  • The further away the point of observation is from the charge, the weaker the field becomes, as it diminishes with the square of the distance.
Point Charge
A point charge is a charge concentrated at a single point in space. This concept is an idealized model used to simplify calculations in electrostatics.
Such charges can represent small charged bodies where their size is negligible compared to the distance from the point of interest.

The charge in the original exercise is \(-3.00 \, \text{nC}\) (nanocoulombs, where \(1 \, \text{nC} = 10^{-9} \, \text{C}\)). When converting units,
\(-3.00 \, \text{nC}\) becomes \(-3.00 \times 10^{-9} \, \text{C}\).
  • Being negative, this charge will create an electric field that attracts positive charges towards it.
  • The strength of this electric field can be calculated using Coulomb's Law,
  • which helps in quantifying the effect of a point charge at various distances.
Electric Field Direction
The direction of the electric field is an essential concept when studying electric fields produced by point charges. Given a negative point charge, like the one in our exercise, the electric field lines point toward the charge.
This is because the electric field direction indicates how a positive test charge would move under the influence of the field.

For example, with our negative charge of \(-3.00 \, \text{nC}\), the electric field direction at any point around it is inward, toward the charge.
  • This contrasts with a positive point charge, where the field would radiate outward.
  • Being able to predict and determine these directions is vital for understanding interactions in electrostatic environments.

Knowing the field direction plays a crucial role in applications, from calculating forces exerted on other charges to understanding electric field distribution in space.

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Most popular questions from this chapter

BIO Electrophoresis. Electrophoresis is a process used by biologists to separate different biological molecules (such as proteins) from each other according to their ratio of charge to size. The materials to be separated are in a viscous solution that produces a drag force \(F_{\mathrm{D}}\) proportional to the size and speed of the molecule. We can express this relation- ship as \(F_{\mathrm{D}}=K R v,\) where \(R\) is the radius of the molecule (modeled as being spherical), \(v\) is its its speed, and \(K\) is a constant that depends on the viscosity of the solution. The solution is placed in an external electric field \(E\) so that the electric force on a particle of charge \(q\) is \(F=q E\) . (a) Show that when the electric field is adjusted so that the two forces (electric and viscous drag) just balance, the ratio of \(q\) to \(R\) is \(K v / E\) . (b) Show that if we leave the electric field on for a time \(T,\) the distance \(x\) that the molecule moves during that time is \(x=(E T / k)(q / R)\) . (c) Suppose you have a sample containing three different biological molecules for which the molecular ratio \(q / R\) for material 2 is twice that of material 1 and the ratio for material 3 is three times that of material 1. Show that the distances migrated by these molecules after the same amount of time are \(x_{2}=2 x_{1}\) and \(x_{3}=3 x_{1}\) . In other words, material 2 travels twice as far as material \(1,\) and material 3 travels three times as far as material \(1 .\) Therefore, we have separated these molecules according to their ratio of charge to size. In practice, this process can be carried out in a special gel or paper, along which the biological molecules migrate. (Fig. P21.94). The process can be rather slow, requiring several hours for separations of just a centimeter or so.

LA A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{N} / \mathrm{C}\) . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00\(\mu\) in the field, assuming it starts from rest.

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 \(\mathrm{mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

A charge of \(-6.50 \mathrm{nC}\) is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 \(\mathrm{cm}\) . (a) Find the magnitude and direction of the electric field this disk produces at a point \(P\) on the axis of the disk a distance of 2.00 \(\mathrm{cm}\) from its center. (b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point \(P\) . (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point \(P .\) (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?
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