Solution MC Hwk (8) - MC HW 8 PDF

Preview

Summary

MC HW 8...

Description

Solution for Multiple-Choice Homework 8 Computing the Fields of Conductors and Dielectrics Solution to Multiple-Choice Homework Problem 8.1(Field for Spherical System) Problem: A conducting spherical shell has an inner radius a and an outer radius b. The shell has a net charge of −3.0nC. A point charge of charge 3.0nC is placed at the center of this shell. Select the figure below that correctly presents the field map for the system: Select One of the Following: (a-Answer) Figure (a)

(b) Figure (b)

(c) Figure (c)

(d) Figure (d)

(e) Figure (e)

(f) Figure

(f)

Figure (b)

Figure (a)

Figure (c)

+ + _Conductor + _ Air

Conductor

_Conductor_ Air

Air +

_

+ _

_

_

+

+ +

Figure (d) _ Conductor _ _ _

_ _

Air

Figure (e) _ Conductor _ + +

Figure (f) + _Conductor + _

Air

_ _

_

+

+

_ _

+

Air

_ +

Solution 4 lines per Q. Use Gauss’ Law to determine the flux (number . of field lines) through each region, φ = Qenc ε0 Region I: Qenc = Q = 4 lines, so there needs to be 4 lines directed from the charge outward in region I. Region III: Qenc = 0, so there should be zero field lines entering/exiting region III. Region II: The electric field in the conductor is zero, so region II has zero field lines.

1

_

_

_

_

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.2(Spherical Total Induced Charge Inner Surface Dielectric) Problem: The figure to the right shows a spherical system consisting of a point charge centered at the origin with total charge +Q and a neutral dielectric shell. The dielectric constant of the shell is κ = 4. What is the total bound charge on the inner surface of the neutral dielectric shell?

dielectric

Select One of the Following: (a) 0

+Q

(b) −Q/4 (c) −Q/2 (d-Answer) −3Q/4 (e) −Q

Solution The electric field that would exist if the dielectric were not there is ~E = charge. The dielectric reduces the field by κ to ~= E

Q rˆ, 4πε0 r 2

the electric field of a point

Q rˆ 4πκε0 r2

This field can also be calculated using a Gaussian surface in the dielectric which encloses a total charge of Qenc = Q + Qbin where Qbin is the bound charge on the inner surface of the dielectric. This produces a field of Q + Qbin E~ = rˆ 4πε0 r2 Equating the two forms of the field gives Q ~ = Q + Qbin rˆ = E rˆ 2 4πε0 r 4πκε0 r2 Canceling gives Q + Qbin = or Qbin =

Q κ

Q 3 Q −Q =− Q −Q = 4 κ 4

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 7.9(Thin Field Lines in Dielectric) Problem: A dielectric with dielectric constant κ is added to a field map. Before the addition of the dielectric, f , field line pass through the space that will be occupied by the dielectric. Approximately how many field lines pass through the dielectric after it is placed in the field? Select One of the Following: (a) 0 (b) f (c) κf 2

(d-Answer) f /κ (e) f /2 Solution The factor by which an electric field is reduced inside a dielectric is called the dielectric constant, denoted κ. Since there were f field lines in the region before the dielectric is present, the number of field lines that would pass through the dielectric if it were placed in this region is f /κ. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.4(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)

Figure (a)

Figure (b) ++

+

+

+

+

+ _

_ _

_

++

++

+

_

_ +

+

_

_

+

+

+

+ +

+ Figure (c) + _ +

_

_ _ +

3

+

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 II.

IV +

+ _

III II I +

+

_

_

+

_ + +

+

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.5(Non-Uniform Spherical System) Problem: A spherical system of charge has NON-UNIFORM volume charge density γ ρ= 4 r and occupies the region from r = a to r = b as drawn to the right where γ is a constant. Compute the electric field at all points OUTSIDE the sphere (r > b).

volume charge a b Air

Select One of the Following: γ rˆ 3ε0 r3 4 πγb3 − 43 πγa3 (b) ~E = 3 rˆ 4πε0 r6 4 πγb3 − 43 πγa3 (c) ~E = 3 rˆ 4πε0 r2 ρ rˆ (d) ~E = 4πε0 r2 ¶ µ 1 1 r ~ = γ (e-Answer) E ˆ − b ε0 r2 a ~ = (a) E

Air

Solution A Gaussian surface of radius r > a encloses a total charge Z a Z Qenc = 4πr2 ρ(r)dr =

γ dr r4 µ ¶¯ Z b 1 ¯a 1 ¯ dr = 4πγ − = 4πγ 2 r ¯b a r µ ¶ 1 1 = 4πγ − a b 0

Qenc

a

4πr2

0

For spherical symmetry, Gauss’ Law becomes µ ¶ 1 1 1 ~ = Qenc rˆ = · 4πγ − rˆ E 4πε0 r2 a 4πε0 r2 b 4

~ = E

¶ γ µ1 1 − rˆ ε0 r2 a b

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.6(Electric Field at Conductor Surface) Problem: Suppose the electric field was not perpendicular to the surface of a conductor, what would happen? Select One of the Following: (a) Nothing would happen if the field was not perpendicular to the conductor’s surface. (b) The electric field would be larger inside the conductor than outside the conductor. (c-Answer) An electric current would flow along the surface of the conductor. (d) The surface atoms of the conductor would become insulating. (e) The resistance of the conductor would increase. Solution If the electric field was not perpendicular to the surface of a conductor, there would be a component of the electric field along the surface and a surface current would flow. 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. 5

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.8(Grounding Conductor Enclosing Fixed Charge) Problem: Two +Q point charges are fixed in a hollow cavity in a conductor as drawn to the right. If the conductor is grounded, what is the net charge of the conductor? Do not include the charge of the two fixed point charges in the charge of the conductor.

conductor

Select One of the Following: (a) 0

+Q

(b) +Q

+Q

(c) −Q (d) +2Q (e-Answer) −2Q

Solution The positive charges in the hollow attract negative charges to the inside surface of the conductor. If the conductor is grounded, then the excess positive charges on the exterior surface of the conductor are allowed to exit, leaving a net negative charge. A Gaussian surface in the conductor must enclose zero charge, so the charge on the inside of the conductor and therefore the charge of the conductor is −2Q. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.9(Surface Charge of Dielectric Half Plane) Problem: A dielectric slab is placed in a external field as shown to the right. Outside the slab, the electric field is ~E0 = (10.0N/C)ˆx. Inside the slab, the electric field is ~Eκ = (3.5N/C)ˆx. Compute the surface charge density on the left side of the slab. Select One of the Following: (a) −1.1 × 10−11 C/m2 (b) −3.1 × 10−11 C/m2 (c) −8.9 × 10−11 C/m2 (d-Answer) −5.8 × 10−11 C/m2 (e) −1.2 × 10−11 C/m2

Solution

6

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

or solving for σ l , σ l = ǫ0 (Eκ −E0 ) = (8.85×10−12

C2 N N )((3.5 )−(10 )) = −5.8×10−11 C/m2 C C Nm2

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 8.10(Draw + Where Lines Begin Where Lines End) Problem: The figure below shows a conductor with zero net charge in an external electric field. Select the figure that correctly represents the surface charge distribution on the conductor. Select One of the Following: (a) Figure (a)

(b) Figure (b)

(c) Figure (c)

(d-Answer) Figure (d)

+ _ + _

Figure (a)

+ _ _ + Figure (b) _ + _ +

+ +

_ + _ +

+ + Figure (c)

Figure (d) Solution

7

In an electric field, a conductor experiences charge separation. Field lines terminate on negative charges and begin on positive charges, so draw − where the field lines end and + where they begin. Also, there must be an equal amount of + and − because the net charge of the conductor is zero.

_ + _ + _ + _ +

Total Points for Problem: 3 Points

8...