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Solution for Multiple-Choice Homework 8 Computing the Fields of Conductors and Dielectrics Solution to Multiple-Choice Homework Problem 8.1(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)

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 8.2(Surface Charge of Dielectric Half Plane) Problem: A sheet of Teflon, an insulator with dielectric constant 2.1, is held so that its surface normal is parallel to the earth’s electric field of 150 NC . Compute the magnitude of the surface charge density on one side of the sheet. Select One of the Following: (a) 1.5 × 10−9 C/m2 (b) 4.1 × 10−9 C/m2 (c) 8.7 × 10−9 C/m2 (d) 3.3 × 10−10 C/m2 (e-Answer) 7.0 × 10−10 C/m2 1

Solution Electric field lines can only end on negative charge. Since the direction of the Earth’s electric field is down and ends on the top of the dielectric, there must be a negative bound charge density on the top surface of the dielectric (see figure). Note that the bottom surface of the dielectric has a positive surface charge density because field lines begin. 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 top surface will then be Q = σ t A. The top’s surface normal is directed upward, which is opposite the Earth’s electric field, so the flux through the top of the pill box is φ = −E0 A. Inside the dielectric, the normal vector is in the same direction as the field, so the flux is φ = +Eκ A. Applying Gauss’ Law, Eκ A − E0 A =

E

N

N

σt A Qenc = ǫ0 ε0

Recall that the electric field inside a dielectric is reduced by factor κ, 150 N E0 N C Eκ = = = 71 C κ 2.1 Solving for σ t , µ ¶µ ¶ C2 N N σ t = ǫ0 (Eκ −E0 ) = 8.85 × 10−12 ) − (150 (71 ) = −7.0×10−10 C/m2 C C Nm2 Therefore, the magnitude of the surface charge density on each side of the conductor is 7.0 × 10−10 C/m2 . Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.3(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

+ + +

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 . 2

_ _ _ _

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

Solution to Multiple-Choice Homework Problem 8.4(Compute Surface Charge of a Plane) Problem: A uniform plane of charge lies in the y − z plane. x. Below Above the plane, the electric field is ~E+ = −300N/Cˆ x. The field is the plane, the electric field is ~E− = −500N/Cˆ drawn to the right. Note, the system must be part of a larger system since the outer field is not equal and opposite. Compute the surface charge density in the y − z plane.

x

E+=300N/C

Select One of the Following:

y

(a) −3.5 × 10−9 mC2 C m2 (c) −7.1 × 10−9 mC2 (d-Answer) 1.8 × 10−9 mC2

(b) 6.4 × 10−9

(e) 0

E−=500N/C Solution

3

(a) Figure Out the Sign from the Field Map: lines begin on the surface, σ > 0.

Since more

ntop

E+

σ nbottom E−

(b) Draw Gaussian Pillbox: The dashed line in the figure is the Gaussian Pillbox, with top and bottom area A. Draw it so the top and bottom of the cylinder are very close to the surface. (c) Argue the Sides Away: Since we make the Gaussian pillbox infinitely short, the flux through the sides is zero. (d) Compute φtop : Use definition of electric flux: Z ~+ · n φtop = (E ˆ top )dA S

~ ˆ top is constant over the top of the Gaussian surface, it can be removed from the integral leaving RSince E+ · n dA = A the area of the top, so C ˆ top A φtop = E~+ · n by observation n ˆ top = x ˆ so

N N φtop = (−300 x ˆ) · (ˆ x)A = (−300 )A C C

(e) Compute φbottom : Likewise ~− · n ˆ bottom A. φbottom = E By observation n ˆ bottom = −ˆ x so φbottom = (−500

N N x ˆ) · (−x)A ˆ = (500 )A C C

(f) Use Gauss’ Law: Gauss’ Law states the flux leaving a closed surface is proportional to the charge enclosed: φtop + φbottom =

Q . ε0

By definition of surface charge, Q = σA where σ is the surface charge of the plane, so σA N N (−300 )A + (500 )A = C C ε0 C2 N N σ = ε0 (200 ) = (8.85 × 10−12 )(200 ) C C Nm2 C σ = 1.8 × 10−9 2 m Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.5(Charge in the Cavity of a Arbitrary Conductor ) 4

Problem: A positive point charge with charge +Q is placed and held inside a cavity in a neutral conductor as drawn. No charge is transferred to the conductor. What is the total charge on the surface of the cavity? No charge is transferred to the environment. Select One of the Following: (a) 0

conductor

(b) +Q

+Q

(c-Answer) −Q (d) negative but not quite −Q (e) positive but not quite +Q

Solution If there is zero electric field inside the conductor, a Gaussian surface drawn within the conductor will enclose zero net charge. Therefore, an amount of charge equal in magnitude and opposite in sign to that in the cavity will move to the conductor’s cavity walls. Note that the total charge of the conductor (excluding the point charge) remains zero because positive charge moves to the exterior surface. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.6(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 8.7(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? Select One of the Following: 5

(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 8.8(Surface Charge Density on a Conductor)

+σ

-σ

Conductor

Problem: The figure to the right shows two parallel planes with . equal and opposite surface charge densities, ±σ, σ = 0.1 µC m2 Calculate the induced charge density on the left and right surface of the conductor.

Select One of the Following: (a-Answer) σ l = −0.1µC/m2 and σ r = +0.1µC/m2 (b) σ l = +0.1µC/m2 and σ r = −0.1µC/m2 (c) σ l = −0.05µC/m2 and σ r = +0.05µC/m2 (d) σ l = +0.05µC/m2 and σ r = −0.05µC/m2 (e) σ l = 0 and σ r = 0 Solution

6

Draw a good field map of the system, remembering that there are no field lines inside the conductor. The field in Regions I and III are both zero. ~I =E ~ III = 0 E

I

+σ

II

−σ

III

EII A − EI A =

Conductor

Now, using a Gaussian surface between Regions I and II, find the field in Region II. Qencl σA = ǫ0 ǫ0

EII =

σ ǫ0

To find the induced charge density on the left surface of the conductor, draw another Gaussian surface that encloses the left surface. Then use Gauss’ Law again: Econd A − EII A =

x

σℓA ǫ0

Electric field inside a conductor is always zero, ~Econd = 0, so −EII =

σℓ ǫ0

σ σℓ = ǫ0 ǫ0 −σ = σ ℓ

−

We can use the same Gaussian surface technique to find the induced charge density on the right surface, or we can use Conservation of Charge on an uncharged conductor. In this case, the charge on the left surface must be equal and opposite to the charge on the right surface so the conductor as a whole will remain uncharged. Therefore, σ r = −σ ℓ = σ = 0.1µC m2 Total Points for Problem: 4 Points

Solution to Multiple-Choice Homework Problem 8.9(Charge Required to Cancel Field) 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. 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 7

E0

-σ

+σ

E0

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 m C Nm2

Solution to Multiple-Choice Homework Problem 8.10(Sign Bound Charge from Field Map) Problem: The figure to the right shows a dielectric slab in a uniform electric field. Is there a net bound charge density on the LEFT surface of the dielectric? If there is, what is the sign of the bound charge density on the LEFT surface of the dielectric?

Dielectric

Select One of the Following: (a) positive (b-Answer) negative (c) There is no net bound charge density on the left surface of the dielectric.

E0 Solution Since the field lines end on the left surface, the bound charge density is negative. Total Points for Problem: 3 Points

8

Eκ

E0...