Solution for Multiple-Choice Homework 7 Conductors and Dielectrics Solution to Multiple-Choice Homework Problem 7.1(Electric Field at Conductor Surface) conductor is placed placed in an electric field. field. What can be said about the electric electric field lines lines at the surface Problem: A conductor of the conduc conductor? tor? Select One of the Following: (a) Any field line configuration is possible at a conductor’s surface. (b) No field lines can intersect a conductor’s surface. (c) Field lines must be tangent to the conductor’s surface. (d-Answer) Field lines must be perpendicular to the conductor’s surface. (e) Field lines must make a 45 a 45 angle to the conductor’s surface. ◦
Solution The electric electric field is perpendic perpendicular ular to the surface of the condu conductor ctor.. If the elec electric tric field was not perpendicular 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 Points for Problem: Problem: 3 Poi Points nts
Solution to Multiple-Choice Homework Problem 7.2(Field for Spherical System) charge ge of char charge ge +3 +3..0nC is located located at the cen center ter of a con conduc ductin tingg sph spheri erical cal shell. shell. The Problem: A point char conducting shell has inner radius a and outer radius b. Selec Selectt the figure below below that represents represents the field map for the system if a net charge of +3 of +3..0nC 0nC is is added to the spherical conductor. Select One of the Following: (a)) Fi (a Figu gure re (a (a))
(b)) Fi (b Figu gure re (b (b))
(c-A (c -Ans nswe wer) r) Fi Figu gure re (c (c))
1
Figure (a) Conductor _
Figure (b)
+
Conductor
_
Air + _
+
Air
_
+
Figure (c) +
+ + _
_
Air +
+ _ +
_
Conductor + +
Solution Gauss’ law states that the number of electric field lines leaving a surface is proportional to the charge enclosed by that surface. Next, divide the system into three regions. Region I: The total charge in this region is Q enc = Q = +3.0nC; there are 4 electric field lines leaving this region. Region II: The electric field in a conductor is zero, so there should not be any field lines drawn. Region III: The total charge enclosed in this region is Qenc = 2Q = +6.0nC. Since region I defined 4 field lines leaving a charge +3.0nC, then there must be 8 field lines for region III.
+ + -
-
III II
I +
+ Air
-
-
+
+ +
2
+
Conductor
Total Points for Problem: 3 Points
Solution to Multiple-Choice Homework Problem 7.3(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 7.4(Charge Density on Hubcap) Problem: On a construction site, a 2.0m by 3.0m conducting metal sheet lies on the ground. The earth produces an electric field of E = 150 N downward. What is the surface charge density C on the upper surface of the metal sheet? Select One of the Following: (a) 2.6 × 10
9
−
(b) −4.7 × 10
C/m2 8
−
C/m2
(c-Answer) −1.3 × 10 (d) −7.8 × 10 (e) 5.1 × 10
7
−
9
−
9
−
Sheet C/m2
C/m2
Earth
C/m2
Solution Using a cylindrical Gaussian surface with one end in the field and one end in the metal sheet, Gauss’ law states φe = Qε . The charge enclosed in the surface is σ A, so we can write enc
0
φe =
σA ε0
Since the field lines enter the Gaussian surface, the flux is negative: φe = −|E |A. Putting it all together and solving for the charge density gives, σ = −ε0 |E | = −(8.85×10
12
−
C2 N )(150 ) = −1.3×10 2 Nm C
9
−
3
C/m2
+
+
-
-
-
+
+
+
Sheet
Earth
Total Points for Problem: 3 Points
Solution to Multiple-Choice Homework Problem 7.5(Compute Bound Charge of Plexiglass) Problem: A piece of glass (κ = 7) is placed above an infinite plane of charge with negative surface charge density σ = −1.5 × 10 7 mC as shown to the right. Compute the bound charge density, σb , on the bottom surface of the glass. Be careful of the sign. −
2
σt σb
Select One of the Following: (a) −6.4 × 10
8
−
−
(c) −3.8 × 10 (d) 3.8 × 10
12
−
12
−
(e) −5.5 × 10
8
-
-
-
-
-
-
C/m2
C/m2
C/m2
6
−
-
C/m2
(b-Answer) 6.4 × 10
κ=7
C/m2 Solution
(a) Calculate the Fields: Outside the slab, the field is that of an infinite plane. So immediately below the slab the field is 0 = − σ zˆ E 2ε0 The field inside the slab is reduced by a factor of κ, κ = − σ zˆ E 2ε0 κ
z σt
σb
-
-
-
-
κ=7
+
-
-
+
+
+
-
-
-
-
(b) Apply Gauss’ Law: Use a Gaussian Pillbox, a cylinder with end area A as drawn above. The charge enclosed in the cylinder is Q enc = σ b A. Applying Gauss’ Law to the pillbox yields E t A − E b A = E κ A − E 0 A =
Qenc σb A = ε0 ε0
(c) Substitute the Fields: Cancelling A and substituting the field,
σ − 2ε0 κ
−
σ σb − = 2ε0 ε0
σ σ σ σb = − + = 2κ 2 2 0.15 µmC σb = 2 2
1 − + 1 κ
1 − + 1 = 6.4 × 10 7 4
8
−
C/m2
Total Points for Problem: 3 Points
Solution to Multiple-Choice Homework Problem 7.6(Integrating a Volume Charge) Problem: A spherically symmetric charge density occupies the region r < a. The charge density increases quadratically with radius as ρ(r) = γ r2 where γ is a constant. Calculate the total charge. Select One of the Following: (a) 2γa γa 3 3 πγa5 (c) 5 (d) 4πγa5 (b)
(e-Answer)
4πγa5 5 Solution
(a) Spherically Symmetry Charge Density: The total charge of a spherically symmetric volume charge density is a Q = 4πr 2 ρ(r)dr
0
This is found by dividing the volume up into this shells of volume 4πr 2 dr. (b) Substitute: Substitute the charge density and set the limits of integration a
Q =
a
2
2
4πr (γr )dr = 4πγ
0
r4 dr
0
(c) Do the Integral: The integration yields r 5 /5 and after imposing the limits of integration Q =
4πγa5 5
Total Points for Problem: 3 Points
Solution to Multiple-Choice Homework Problem 7.7(Surface Charge Density on a Conductor) Problem: The figure to the right shows two parallel planes with equal and opposite surface charge densities, ±σ, σ = 0.1 µmC . Calculate the induced charge density on the left and right surface of the conductor. 2
+σ
-σ
r o t c u d n o C
Select One of the Following:
5
(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 Draw a good field map of the system, remembering that there are no field lines inside the conductor. The field in Regions I and I II are both zero. I = E II I = 0 E Now, using a Gaussian surface between Regions I and II, find the field in Region II. Qencl σA E II A − E I A = = ǫ0 ǫ0 σ E II = 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: E cond A − E II A =
I
II
+σ
+
-
+
-
+
-
+
-
−σ
+ r o + t c u d n o C+
+
III
-
x
σℓ A ǫ0
cond = 0, so Electric field inside a conductor is always zero, E −E II =
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, σr = −σℓ = σ = 0.1µC m2
Total Points for Problem: 4 Points
Solution to Multiple-Choice Homework Problem 7.8(Surface Charge Density on Conductor)
6
A spherical conductor has net positive charge as Problem: 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 7.9(Compute Surface Charge of a Plane) + = 10N/Cˆz. Problem: A uniform surface charge lies in the x − y plane. Above the plane, the electric field is E = 3N/Cˆz. Note, the system must be part of a larger system since the Below the plane, the electric field is E outer field is not equal and opposite. Compute the surface charge density of the plane. −
Select One of the Following: (a-Answer) 6 × 10
11 C m2
−
(b) 1 × 10
−
10 C m2
(c) −6 × 10
11 C m2
(d) −1 × 10
10 C m2
−
−
(e) 0 Solution
7
Definitions E +
z
+ = 10 N ˆz ≡ Electric Field above Plane E C N = 3 ˆz ≡ Electric Field below Plane E C n ˆ bottom ≡ Normal vector of Bottom of Gaussian Surface n ˆ top ≡ Normal vector of Top of Gaussian Surface
ntop
−
σ
σ ≡ Charge Density of plane φtop ≡ Flux through top of Gaussian Surface φbottom ≡ Flux through bottom of Gaussian Surface A ≡ Area of end of Gaussian Surface
nbottom
E −
Strategy: Draw an infinitely thin cylindrical Gaussian Surface, and use Gauss’ law to compute charge inside. (a) Draw a Good Diagram and Field Map: Draw the field map using the general techniques of next section. Since more lines begin on the surface, σ > 0. (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: φtop =
(E + · n ˆ top )dA
S
+ · ˆntop is constant over the top of the Gaussian surface, it can be removed from the integral leaving Since E dA = A the area of the top, so C + · n φtop = E ˆ top A
by observation n ˆ top = zˆ so
N N φtop = (10 ˆz) · (ˆz)A = (10 )A C C
(e) Compute φbottom : Likewise · n φbottom = E ˆ bottom A. −
ˆ bottom = −ˆ By observation n z so
N N φbottom = (3 ˆz) · (−ˆz)A = (−3 )A C C (f) Use Gauss’ Law: Gauss’ Law states the flux leaving a closed surface is proportional to the charge enclosed: Q . ε0
φtop + φbottom =
By definition of surface charge, Q = σA where σ is the surface charge of the plane, so N N σA (10 )A + (−3 )A = C C ε0 N σ = ε 0 (7 ) = (8.85 × 10 C σ = 6 × 10
Total Points for Problem: 3 Points
8
11
−
12
−
C m2
C2 N )(7 ) 2 Nm C
Solution to Multiple-Choice Homework Problem 7.10(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 N . Compute the magnitude of the surface charge density on one side of the C 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 10
(e-Answer) 7.0 × 10
−
C/m2 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 = −E 0 A. Inside the dielectric, the normal vector is in the same direction as the field, so the flux is φ = +E κ A. Applying Gauss’ Law, E κ A − E 0 A =
E
N
N
Qenc σt A = ε0 ǫ0
Recall that the electric field inside a dielectric is reduced by factor κ, 150 N E 0 N C E κ = = = 71 κ 2.1 C Solving for σ t , σt = ǫ0 (E κ −E 0 ) =
8.85 × 10
−
12
C2 Nm2
N N (71 ) − (150 ) = −7.0×10 C C
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
9
10
−
C/m2
