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Chapter 22: Problem 50

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

Short Answer

Expert verified
(a) \( n \approx 6.73 \times 10^{11} \) excess electrons. (b) \( E \approx 414 \) N/C.

Step by step solution

01

Understanding the Problem

We need to calculate the number of excess electrons needed to create a specified electric field just outside a charged sphere. The diameter of the sphere is 30 cm, which means the radius is 15 cm. The electric field just outside the surface is given as 1150 N/C.
02

Calculating Charge Required

The electric field just outside a uniformly charged sphere is given by the formula \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \) N路m虏/C虏, \( Q \) is the charge, and \( r \) is the radius of the sphere. Rearrange to solve for \( Q \): \( Q = \frac{Er^2}{k} \). Substitute \( E = 1150 \) N/C and \( r = 0.15 \) m to find \( Q \).
03

Substituting Values

Substitute the values into the formula for \( Q \): \[ Q = \frac{1150 \times (0.15)^2}{8.99 \times 10^9} \]. Calculate \( Q \) to find the charge in coulombs.
04

Converting Charge to Electrons

An electron has a charge of \( -1.6 \times 10^{-19} \) C. To find the number of electrons (\( n \)) that corresponds to the charge \( Q \), use \( n = \frac{Q}{1.6 \times 10^{-19}} \). Substitute \( Q \) from the previous step to find the value of \( n \).
05

Calculating Electric Field at a Distance

For part (b), use \( E = \frac{kQ}{R^2} \) where \( R \) is the total distance from the sphere's center to the point of interest, which includes the sphere's radius plus 10 cm. Thus, \( R = 0.25 \) m. Substitute \( Q \) from part (a) and \( R \) into the formula to calculate \( E \).

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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 fundamental concept in electrostatics. It's a region around a charged object where other charges experience a force. For a charged sphere, the electric field just outside its surface behaves as if all the charge were concentrated at the center. This allows us to simplify calculations.
An electric field strength is expressed in newtons per coulomb (N/C). In this exercise, the electric field strength just outside the sphere is given as 1150 N/C, which helps us determine how many excess electrons need to be distributed on the sphere.
Coulomb's Law
Coulomb's Law provides a quantitative description of the electric force between charged objects. The law states that the force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The formula used is:
  • \( F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \)
Where:
  • \( k \) is Coulomb's constant, valued at \( 8.99 \times 10^9 \) N路m虏/C虏.
  • \( Q_1 \) and \( Q_2 \) are the charges.
  • \( r \) is the distance between the charges.
While this exercise doesn't compute forces, it uses the principles of Coulomb's Law to understand electric fields' dependencies on charge and distance.
Charge Distribution
Charge distribution tells us how charge is spread over or within an object. In this scenario, an isolated plastic sphere is assumed to have a uniform charge distribution, making the electric field calculation straightforward. The charge is spread evenly over the sphere's surface, allowing the entire charge to be treated as if it were concentrated at the center for the purpose of calculating the electric field outside the sphere.
Uniform distribution simplifies many calculations and helps us use formulas involving radial distribution, crucial for this step-by-step solution.
Excess Electrons
Excess electrons are additional electrons that create a net negative charge on an object. Determining the number of excess electrons involves calculating the net charge and then converting this charge to the number of electrons, using each electron's charge value of \(-1.6 \times 10^{-19}\) C.
After calculating the required charge to make the electric field just outside the sphere 1150 N/C, the number of electrons is given by:
  • \( n = \frac{Q}{1.6 \times 10^{-19}} \)
This calculation allows us to quantify exactly how much charge, in terms of electron numbers, is necessary to produce the desired electric field around the sphere.

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Most popular questions from this chapter

22.24. A point charge of \(-2.00 \mu \mathrm{C}\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3} .\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

22.58. A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho(r)\) given as follows: $$ \begin{array}{ll}{\rho(r)=\rho_{0}(1-4 r / 3 R)} & {\text { for } r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array} $$ where \(\rho_{0}\) is a positive constant. (a) Find the total charge contained in the charge distribution. (b) Obtain an expression for the electric field in the region \(r \geq R .\) (c) Obtain an expression for the electric field in the region \(r \leq R .(d)\) Graph the electric-field magnitude \(E\) as a function of \(r .(e)\) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field. $$ \oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G m $$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show the flux of \(\overrightarrow{\boldsymbol{g}}\) through any closed surface is given by $$ \oint \overrightarrow{\boldsymbol{g}} \cdot d \boldsymbol{A}=-4 \pi G M_{\mathrm{encl}} $$

22\. 13. A \(9.60-\mu\) C point charge is at the center of a cube with sides of length 0.500 \(\mathrm{m}\) (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 \(\mathrm{m}\) long? Explain.

22.63. Positive charge \(Q\) is distributed uniformly over each of two spherical volumes with radius \(R\) . One sphere of charge is centered at the origin and the other at \(x=2 R\) (Fig. 2244 ). Find the magnitude and direction of the net electric field due to these two distributions of charge at the following points on the \(x\) -axis: (a) \(x=0 ;\) (b) \(x=R / 2 ;(c) x=R ;\) (d) \(x=3 R\) .

22\. 28 . A square insulating sheet 80.0 \(\mathrm{cm}\) on a side is held horizontally. The sheet has 7.50 \(\mathrm{nC}\) of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 \(\mathrm{mm}\) above the center of the sheet. (b) Estimate the electric field at a point 100 \(\mathrm{m}\) above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?
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