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Chapter 17: Problem 59

\(\bullet\)(a) How many excess elec- trons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an elec- tric field of 1150 \(\mathrm{N} / \mathrm{C}\) just out- side the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

Short Answer

Expert verified
(a) Approximately \(1.802 \times 10^{10}\) excess electrons are needed. (b) The electric field 10 cm outside is approximately 415.9 N/C.

Step by step solution

01

Understand the Problem

We need to determine the number of excess electrons needed on the surface of a sphere to produce a given electric field at its surface and then find the electric field at a point outside the sphere.
02

Use Gauss's Law for Electric Field

For a sphere of radius \( R = 15.0 \text{ cm} = 0.15 \text{ m} \), by Gauss's Law, the electric field \( E \) just outside the surface of the sphere is given by \( E = \frac{kQ}{R^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is the Coulomb's constant, and \( Q \) is the total charge on the sphere.
03

Calculate Total Charge Required

Given \( E = 1150 \text{ N/C} \), solve for \( Q \):\[ Q = \frac{E R^2}{k} = \frac{1150 \times (0.15)^2}{8.99 \times 10^9} \approx 2.887 \times 10^{-9} \text{ C} \]
04

Determine Number of Excess Electrons

Each electron has charge \( e = 1.602 \times 10^{-19} \text{ C} \). The number of excess electrons \( n \) is given by:\[ n = \frac{Q}{e} = \frac{2.887 \times 10^{-9}}{1.602 \times 10^{-19}} \approx 1.802 \times 10^{10} \]
05

Calculate Electric Field at a Point Outside the Sphere

The electric field at a distance \( r = 0.15 + 0.10 = 0.25 \text{ m} \), outside the sphere can be calculated using the formula:\[ E' = \frac{kQ}{r^2} = \frac{8.99 \times 10^9 \times 2.887 \times 10^{-9}}{(0.25)^2} \approx 415.9 \text{ N/C} \]

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Key Concepts

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

Electric Fields
An electric field is a vector field around charged particles. It represents the force that a charged particle would experience when placed within the field. Electric fields are crucial in understanding electrical interactions, as they describe how charges influence one another by exerting forces from a distance.

In this exercise, the electric field is given as 1150 N/C at the surface of a sphere, requiring us to calculate the charge distributed over the sphere's surface. Once the charge is determined, it describes the level of force that would act on another charge placed just outside this sphere.

Key points to remember about electric fields include:
  • The direction of the field is away from positive charges and towards negative charges.
  • It is measured in newtons per coulomb (N/C).
  • In solutions, it's common to use Gauss's Law to relate the electric field to the charge enclosure.
Spheres
Spheres, in physics, often refer to geometrical objects with uniform symmetry all over their surface. This makes them ideal candidates for using Gauss鈥檚 law when calculating electric fields. In our problem, the sphere is made of plastic and is used to hold excess electrons, creating an electric field around it.

When dealing with spheres in the context of electric fields:
  • The radius is crucial as it determines the field's spread and strength outside the surface.
  • Electric fields outside a conductive material sphere behave similarly to those generated by a point charge located at the sphere's center.
  • In this exercise, the sphere is isolated, implying no external influences from other charges or fields.
This understanding helps in solving situations where symmetry simplifies calculations, particularly in spherical charge distributions.
Excess Electrons
Excess electrons on a surface create an electric field due to their negative charge. In the exercise, we're determining the number of excess electrons needed to achieve a specific electric field strength.

To find the number of excess electrons:
  • Calculate the total charge needed using the electric field and Gauss鈥檚 law.
  • Divide the total charge by the charge of a single electron, which is approximately \(1.602 \times 10^{-19}\) C.
Excess electrons concentrate the negative charge on the sphere, reinforcing the sphere's capacity to impact the electric field around it. This understanding is essential in applications like capacitance where charge distribution affects performance.
Coulomb's Constant
Coulomb's constant, denoted as \(k\), is a proportionality factor in Coulomb's law of electrostatics, used to calculate the electric force between two point charges. Its value is approximately \(8.99 \times 10^9 \text{ N m}^2/ ext{C}^2\). Understanding this constant is key to solving many electrostatic problems, including our sphere scenario.

Here's why it's important:
  • It provides the scaling factor for electric force and field magnitude calculations.
  • Central to the formula used in Gauss's law \(E = \frac{kQ}{r^2}\), which describes the sphere's surface electric field strength.
Using Coulomb鈥檚 constant, we accurately calculate the charge needed on the sphere to produce the electric field and understand the strength and integrity of electrical interactions in various mediums.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

\(\bullet\) A positively charged rubber rod is moved close to a neutral copper ball that is resting on a nonconducting sheet of plastic. (a) Sketch the distribution of charges on the ball. (b) With the rod still close to the ball, a metal wire is briefly connected from the ball to the earth and then removed. After the rubber rod is also removed, sketch the distribution of charges (if any) on the copper ball.

\(\bullet$$\bullet\) Two point charges are placed on the \(x\) axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) What are the magnitude and direction of the net force exerted by these two charges on a negative point charge \(q_{3}=-0.600 \mathrm{nC}\) placed at the origin?

\(\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.

\(\bullet\) \(\bullet\) The electric field due to a certain point charge has a magni- tude \(E\) at a distance of 1.0 \(\mathrm{cm}\) from the charge. (a) What will be the magnitude of this field (in terms of \(E\) ) if we move 1.0 \(\mathrm{cm}\) farther away from the charge? (b) What will be the magnitude of the field (in terms of \(E )\) if we move an additional 1.0 \(\mathrm{cm}\) far- ther away than in part (a)?
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