Onlinehomework-8-sln PDF

Preview

Summary

Onlinehomework-8-sln...

Description

Solution for Multiple-Choice Homework 8 Computing the Fields of Conductors and Dielectrics Solution to Multiple-Choice Homework Problem 8.1(Why Zero Force in Conductor?) Problem: A short time after a conductor is placed in an external electric field the net electric force felt by the mobile charge inside the conductor becomes zero. What explains this effect physically? Select One of the Following: (a) Mobile charges block the field lines. (b-Answer) Charge separates in response to the external field and sets up a field that cancels the external field in the conductor’s interior. (c) The surface atoms lose electrons to the air to block the field. (d) The atoms in the conductor absorb the electric field. (e) A Gaussian surface enclosing the conductor contains no net charge. Solution Charge separated in response to the external field and set up a field that canceled the external field in the conductors interior. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.2(Surface Charge of Dielectric Half Plane) Problem: An external field, as shown to the right, has a dielectric slab of κ = 3.2 inserted into it. Outside the slab, the electric ~ 0 = (12.0N/C)ˆx. Compute the surface charge density field is E on the left side of the slab. Select One of the Following: (a) −1.1 × 10−11 C/m2 (b) −2.3 × 10−11 C/m2 (c-Answer) −7.3 × 10−11 C/m2 (d) −5.5 × 10−11 C/m2 (e) −8.9 × 10−11 C/m2

Solution

1

Electric field lines can only end on negative charge. Since there is a bigger positively directed field in the region x < 0 than x > 0, there must be a negative bound charge density on the surface of the dielectric (see figure). We can apply Gauss’ Law to the Gaussian surface enclosing the surface of the dielectric to determine the amount of charge on the surface. We choose a Gaussian pillbox with cross sectional area A, since it has the correct symmetry for the planar geometry. The charge enclosed on the left surface will then be σ l A. Applying Gauss’ Law, Eκ A − E0 A =

σl A ǫ0

Recall that the electric field inside a dielectric is reduced by factor κ, E0 N Eκ = = 3.8 C κ Solving for σ l , σ l = ǫ0 (Eκ −E0 ) = (8.85×10−12

C2 N N )((3.8 )−(12 )) = −7.3×10−11 C/m2 2 C Nm C

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.3(Electric Field For a System of Conductors) Problem: A solid spherical conductor with total charge −Q/2 is surrounded by a spherical conducting shell with charge +Q. Select the figure below that correctly represents the field map for the system. Select One of the Following: (a) Figure (a)

(b-Answer) Figure (b)

(c) Figure (c)

2

Figure (a)

Figure (b) ++

+

+

+

+

+ _

_ _

_

++

++

_

_

+

_ +

+

+

_ +

+

+ +

+ Figure (c) + _ +

_

_

+

_ +

Solution First, the system must be divided into four regions to be enclosed by Gaussian surfaces. The first region labeled I is inside a conductor, and therefore has zero electric field. The region labeled II encloses the surface of the inner conductor and all of the charge on the surface. Gauss’ law states that the number of electric field lines leaving a surface is directly proportional to the charge enclosed by the surface. Since there is a net charge enclosed by the Gaussian surface, there must be some electric field lines entering or exiting the surface. The direction of the field lines is determined by the net charge of the inner surface: it has a net negative charge, so field lines point to the inner surface. The electric field in region III is zero due to the conductor. In region IV , the total charge enclosed by the Gaussian surface is −Q/2 + Q = Q/2, so there is an electric field with lines pointing away from the conductor, and the number of field lines in region IV is equal to the number of field lines in region I I. Total Points for Problem: 3 Points

3

IV +

+ _

III II I +

+

_

_

+

_ +

+

+

Solution to Multiple-Choice Homework Problem 8.4(Charge in the Hollow of a Conducting Sphere) Problem: The figure to the right shows a point charge of charge +Q suspended in a cavity of a neutral conductor. The point charge is not at the center of the spherical cavity. Which of the following best describes the charge distribution on the surface of the cavity?

conductor

Select One of the Following: (a) The charge density is zero.

+Q

(b) The charge density is uniform. (c-Answer) The charge density is non-uniform. (d) The charge density is radial. (e) The charge density is perpendicular.

Solution Electrons in the conductor are attracted to the inserted charge more strongly on the side of the cavity the charge is closest to. Thus, the charge distribution on the hollow surface will not be uniform. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.5(Electric Field inside Hollow Conductor Is Zero) Problem: A spherical shell of radius R has uniform surface charge density σ . Three points are labeled inside the sphere. Rank the strength of the electric force, F , on a point charge of charge +Q if it were placed at the various points. Select One of the Following: (a-Answer) FA = FB = FC (b) FA < FB < FC

A

(c) FA > FB > FC

B C

Solution H . Because the surface Consider a Gaussian sphere drawn inside the conductor. Gauss’ Law gives S E · dS = qenc ǫ0 R qenc is symmetric to the field, E dA = ǫ0 . And, because no charge is enclosed, no field is present anywhere inside the shell. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.6(The Meaning of Charge at a Microscopic Level) Problem: In most cases, what causes a region of net positive surface charge in a conductor at the microscopic level? 4

Select One of the Following: (a) Protons have been moved to that region to make a net positive charge. (b) Electrons have been moved to the region to make a net positive charge. (c) Protons have been removed from the region to make a net positive charge (d-Answer) Electrons have been removed from the region to make a net positive charge. (e) Charge is solely a macroscopic phenomenon and has no microscopic origin. Solution The net charge which moves in a conductor is carried by the electrons, so a positively charged region in a material where there are fewer electrons than are required for the material to be electrically neutral. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.7(Surface Charge Density on Conductor) Problem: A spherical conductor has net positive charge as drawn to the right. A negatively charged object with charge −Q is brought near the conductor but does not touch or transfer charge to the conductor. Four points A-D are labelled. At which point or points will the magnitude of the surface charge density of the conductor be greatest?

B

-Q A

positively charged conductor

C

D

Select One of the Following: (a) The charge density will be equal at all points. (b-Answer) The charge density is largest at point A. (c) The charge density is largest at point C . (d) The charge density is largest at points B and D. (e) The charge density is largest in the center of the conductor. Solution Since the negative object will attract more positive charge close to itself, the conductor’s positive surface charge density will be highest closest to the charged object, point A. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.8(Charge Required to Cancel Field)

5

Problem: The figure to the right shows two infinite parallel planes with equal and opposite surface charge densities ±σ. If the external electric field is 100, 000 NC , what magnitude must σ have to reduce the field between the planes to zero. NOTE ~E0 is the EXTERNAL APPLIED FIELD which must be generated by charge densities outside the system.

E0

-σ

+σ

E0

Select One of the Following: (a) 1 × 105 mC2 (b) 1 × 1016

C m2

(c-Answer) 9 × 10−7 mC2 C (d) 6 × 10−8 m 2

(e) 9 × 10−12 mC2 Solution Total Points for Problem: 3 Points The electric field of the two planes must be equal to and opposite of the applied field. The electric field of two equal and opposite planes is |E| = σ/ε0 . Therefore, the charge density needed on the planes is σ = ε0 |E| = (8.85 × 10−12

C2 C N )(100, 000 ) = 9 × 10−7 2 C Nm2 m

Solution to Multiple-Choice Homework Problem 8.9(Zero Lines Leave Neutral Object) Problem: The figure below shows a conductor with zero net charge in an external electric field. Select the figure that could represent the electric field and the neutral conductor. Select One of the Following: (a) Figure (a)

(b-Answer) Figure (b)

(c) Figure (c)

(d) Figure (d)

Figure (a)

Figure (b)

Figure (c)

Figure (d)

6

Solution Since the object is neutral, zero net field lines must begin on the object. If a field line enters the conductor, another must exit because the object is neutral. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.10(Field Lines of Equal and Opposite Planes) Problem: Which of the following best describes the field outside the infinite parallel planes of charge shown to the right? The uniform charge densities on the planes have equal magnitudes, but opposite signs. The system is divided into regions labelled I , II , and III.

_

+

_

+ +

I

II

+ + +

_

III

_ _ _

Select One of the Following: (a-Answer) The field is zero in region I and III . (b) The field points to the left of the page in region I and to the left of the page in region III . (c) The field points to the left of the page in region I and to the right of the page in region III . (d) The field points to the right of the page in region I and to the left of the page in region III . (e) The field points to the right of the page in region I and to the right of the page in region III . Solution The field of equal and opposite planes is zero outside the planes. Imagine a cylindrical Gaussian surface beginning inside region I and ending inside region III. The total charge enclosed is zero (+Q from the left plate and −Q from the right plate), so Gauss’ law says that the flux out of the cylindrical surface must be zero. The argument should be a little more detailed, but we get the idea.

+ +

_

+

_

+ + +

Total Points for Problem: 3 Points

7

_

_ _ _...