HW 06 - Electric Potential Calculations; Capacitance

10/9/2016

HW 06 Electric Potential Calculations; Capacitance

HW 06 Electric Potential Calculations; Capacitance Due: 11:59pm on Sunday, September 25, 2016 To understand how points are awarded, read the Grading Policy for this assignment.

Item 1 A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius 6.00 cm . Both shells are made of insulating material. The smaller shell has charge q1 = + 6.00 nC distributed uniformly over its surface, and the larger shell has charge q2 = − 9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. You may want to review (

pages 765 769) .

For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Potential vsurface and field magnitude esurface at the surface.

Part A What is the electric potential due to the two shells at the following distance from their common center: r = 0? Express your answer with the appropriate units. ANSWER: V

= 0 V

Correct IDENTIFY: The potential at any point is the scalar sum of the potential due to each shell. SET UP: V

kq

=

EXECUTE: (i) r V = k(

q1 R1

for r

R

= 0

+

q2 R2

V = +1.798 × 10

(ii) r

= 4.00cm

V = k(

q1 r

+

V = k( V = 0V

q1 r

cm

+

and V

=

kq r

for r

> R

.

. This point is inside both shells so

) = (8.99 × 10

3

9

−9

2

2

N ⋅ m /C )(

V + (−1.618 × 10

3

V) = 180 V

6.00×10

C

0.0300 m

−9

−9.00×10

C

0.0500 m

.

)

.

. This point is outside shell 1 and inside shell 2. q2 R2

) = (8.99 × 10

V = +1.348 × 10

(iii) r = 6.00

≤ R

3

9

−9

2

2

N ⋅ m /C )(

V + (−1.618 × 10

3

6.00×10

V) = −270 V

0.0400 m

C

−9

−9.00×10

0.0500 m

C

)

.

.

This point is outside both shells. q2 r

9

)=

k r

(q

1

+q )= 2

8.99×10

2

N⋅m /C

0.0600 m

2

[6.00 × 10

−9

C + (−9.00 × 10

−9

C)]

.

.

Part B What is the electric potential due to the two shells at the following distance from their common center: r = 5.00 cm ? Typesetting math: 100% Express your answer with the appropriate units.

https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4769927

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ANSWER: V

= 270 V

Correct At the surface of the inner shell, r V1 = k(

q1 R1

+

q2 R2

smaller shell, so V

) = 180 V

= k(

V2 = +1.079 × 10

3

q1 r

+

. This point is inside the larger shell, so

= R1 = 3.00 cm

. At the surface of the outer shell, r

q2 R2

) = (8.99 × 10

V + (−1.618 × 10

3

9

2

2

N ⋅ m /C )(

V) = −270V

. This point is outside the

= R2 = 5.00 cm −9

6.00×10

C

0.0500 m

−9

−9.00×10

0.0500 m

C

)

.

.

Part C What is the electric potential due to the two shells at the following distance from their common center: r = 7.00 cm ? Express your answer with the appropriate units. ANSWER: V

= 385 V

Correct

Part D What is the magnitude of the potential difference between the surfaces of the two shells? Express your answer with the appropriate units. ANSWER: ΔV

= 450 V

Correct The potential difference is V1

− V2 = 450V

.

Part E Which shell is at higher potential: the inner shell or the outer shell? ANSWER: The inner shell is at higher potential. The outer shell is at higher potential.

Typesetting math: 100%


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Correct The inner shell is at higher potential. The potential difference is due entirely to the charge on the inner shell. EVALUATE: Inside a uniform spherical shell, the electric field is zero so the potential is constant (but not necessarily zero).

Item 2 A solid conducting sphere has net positive charge and radius R = 0.800 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 27.0 V . Assume that V = 0 at an infinite distance from the sphere.

Part A What is the electric potential at the center of the sphere? Express your answer with the appropriate units. ANSWER: V

= 40.5 V

Correct

Item 3 The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments.

Part A What is the work done by the electric force to move a 1 C charge from A to B? Express your answer in joules. Typesetting math: 100%

Hint 1. Find the potential difference between A and B − https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4769927

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What is the potential difference VA

− VB

between point A and point B?

Express your answer in volts.

Hint 1. Equipotential surfaces Recall that an equipotential surface is a surface on which the electric potential is the same at every point. ANSWER: VA − VB

= 0 V

Hint 2. Potential difference and work Recall that the electric potential energy difference between any two points is equal to the negative of the work done by the electric force as a charged object moves between those two points. If we combine this with the relationship between electric potential energy and electric potential we have: WA→B, by electric f orce = −ΔVA→B Q . ANSWER: 0 J

Correct

Part B What is the work done by the electric force to move a 1 C charge from A to D? Express your answer in joules.

Hint 1. Find the potential difference between A and D What is the potential difference VD − VA between point A and point D? Express your answer in volts. ANSWER: VD − VA

= 1 V

Hint 2. Potential difference and work Recall that the electric potential energy difference between any two points is equal to the negative of the work done by the electric force as a charged object moves between those two points. If we combine this with the relationship between electric potential energy and electric potential we have: WAD, by electric f orce = −ΔVAD Q ANSWER: Typesetting math: 100%


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1 J

Correct

Part C The magnitude of the electric field at point C is

Hint 1. Electric field and equipotential surfaces Since the diagram shows equal potential differences between adjacent surfaces, equal amounts of work are done to move a particular charge from one surface to the next adjacent one. It follows then that if the equipotentials are closer together, the electric force does the same amount of work in a smaller displacement than if the equipotentials were farther apart. Therefore, the electric force, as well as the corresponding electric field, has a larger magnitude. ANSWER: greater than the magnitude of the electric field at point B. less than the magnitude of the electric field at point B. equal to the magnitude of the electric field at point B. unknown because the value of the electric potential at point C is unknown.

Correct

Item 4 In a certain region of space, the electric potential is V (x, y, z) constants. You may want to review (

= Axy − Bx

2

+ Cy

, where A, B, and C are positive

pages 771 774) .

For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Potential and field of a point charge.

Part A Calculate the xcomponent of the electric field. Express your answer in terms of the given quantities. ANSWER: Ex

=

−Ay + 2Bx



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Correct IDENTIFY and SET UP: Use Ex EXECUTE: V ∂V

Ex = −

∂x

= Axy − Bx

2

=−

+ Cy

= −Ay + 2Bx

∂V ∂x

, Ey = −

∂V ∂y

, and Ez

=

∂V ∂z

to calculate the components of E⃗ .

.

.

Part B Calculate the ycomponent of the electric field. Express your answer in terms of the given quantities. ANSWER: Ey

=

−Ax − C

Correct Ey = −

∂V ∂y

= −Ax − C

.

Part C Calculate the zcomponent of the electric field. Express your answer in terms of the given quantities. ANSWER: Ez

= 0

Correct Ez = E = 0

∂V ∂z

=0

.

requires that Ex

= Ey = Ez = 0

.

Part D At which point is the electric field equal to zero? ANSWER: x

= 0, y = 0, z = 0

x

= −C/A , y = 0 , z = −2BC/A

x

= −C/A , y = −2BC/A

2

2

, z = C/A



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Correct Ez = 0

everywhere.

Ey = 0

at x

.

= −C/A

And Ex is also equal to zero for this x, any value of z and y EVALUATE: V doesn’t depend on z so Ez

= 0

= 2Bx/A = (2B/A)(−C/A) = −2BC/A

2

.

everywhere.

Item 5 The plates of a parallelplate capacitor are 2.00 mm apart, and each carries a charge of magnitude 83.0 nC . The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×106 V/m . You may want to review (

pages 786 790) .

For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallelplate capacitor.

Part A What is the potential difference between the plates? ANSWER: V

= 1.00×104 V

Correct IDENTIFY: The capacitance depends on the geometry (area and plate separation) of the plates. SET UP: For a parallelplate capacitor, Vab EXECUTE: Vab

= Ed, E =

6

= Ed = (5.00 × 10 V/m)(2.00 × 10

Q ϵ0 A

−3

, and. C =

Q

.

Vab 4

m) = 1.00 × 10 V

.

Part B What is the area of each plate? ANSWER: S

= 1.88×10−3 m2

Correct Solving for the area gives A=

Q Eϵ0

−8

8.30×10

=

6

C −12

(5.00×10 V/m)[8.854× 10

2

= 1.88 × 10

−3

m

2

= 0.19cm

2

.

C /(N⋅m2 )]



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Part C What is the capacitance? ANSWER: C

= 8.30×10−12 F

Correct C =

Q Vab

−8

=

8.30×10

4

C

= 8.30 × 10

−12

F = 8.30pF

.

1.00×10 V

EVALUATE: The capacitance is reasonable for laboratory capacitors, but the area is rather large.

Item 6 A parallelplate air capacitor with a capacitance of 232 pF has a charge of magnitude 0.148 μC on each plate. The plates have a separation of 0.257 mm . You may want to review (

pages 786 790) .

For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallelplate capacitor.

Part A What is the potential difference between the plates? ANSWER: V

= 638 V

Correct IDENTIFY and SET UP: It is a parallelplate air capacitor, so we can apply the equations of Section 24.1. EXECUTE: C =

Q Vab

so Vab

=

Q C

=

0.148μC 232pF

= 638V

.

Part B What is the area of each plate? Use 8.854×10−12 F/m for the permittivity of free space. ANSWER: A

= 6.73×10−3 m2



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Correct C =

ϵ0 A

so A =

d

Cd ϵ0

=

(232pF)(0.257mm) −12

8.854×10

= 6.73 × 10

−3

m

2

= 90.8 cm

2

.

F/m

Part C What is the electric field magnitude between the plates? ANSWER: E

= 2.48×106 V/m

Correct so E =

Vab = Ed

Vab d

=

638V 0.257mm

6

= 2.48 × 10 V/m

.

Part D What is the surfacecharge density on each plate? ANSWER: σ

= 2.20×10−5 C/m2

Correct E=

σ ϵ0

so σ

6

= E ϵ0 = (2.48 × 10 V/m)(8.854 × 10

−12

EVALUATE: We could also calculate σ directly as Q/A. σ = checks.

F/m) = 2.20 × 10

Q A

=

−5

0.148μC −3

9.08×10

C/m

2

.

= 2.20 × 10

−5

C/m

2

, which

2

m

Item 7 A 5.60 μF , parallelplate, air capacitor has a plate separation of 3.90 mm and is charged to a potential difference of 400 V . You may want to review (

pages 794 797) .

For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Electricfield energy.

Part A Calculate the energy density in the region between the plates. ANSWER: u

= 4.65×10−2 J/m3



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Correct IDENTIFY and SET UP: The energy density is given by u = EXECUTE: Calculate E : E = Then u =

1 2

ϵ0 E

2

=

1 2

V d

−3

3.90×10

(8.85 × 10

−12

2

ϵ0 E

2

. Use V

5

400V

=

1

to solve for E .

= Ed

.

= 1.03 × 10 V/m m 5

2

F/m)(1.03 × 10 V/m)

= 4.65 × 10

−2

J/m

3

.

EVALUATE: E is smaller than the value in Example 24.8 by about a factor of 6 so u is smaller by about a factor of 2 6 = 36.

Item 8 A parallelplate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC.

Part A What is the potential difference between the plates? Express your answer with the appropriate units. ANSWER: V

= 4.2 kV

Correct

Part B If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? Express your answer with the appropriate units. ANSWER: V

= 8.5 kV

Correct

Part C How much work is required to double the separation? Express your answer with the appropriate units. ANSWER: ΔU

= 8.3 mJ



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Correct Score Summary: Your score on this assignment is 96.8%. You received 9.68 out of a possible total of 10 points.



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HW 06 - Electric Potential Calculations; Capacitance

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