HW 2-solutions - Quest problems and solutions - Electric field, electric potential PDF

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Quest problems and solutions - Electric field, electric potential...

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bausinger (nlb934) – HW 2 – shih – (55050) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 21 cm long is uniformly charged and has a total charge of −25.7 µC. Find the magnitude of the electric field along the axis of the rod at a point 54.0821 cm from the center of the rod. The Coulomb constant is 8.98755 × 109 N · m2/C2 .

and the magnitude of the electric field is Q E = ke d (ℓ + d)   = 8.98755 × 109 N · m2 /C2 |−2.57 × 10−5 C| × (0.435821 m)(0.21 m + 0.435821 m) = 8.20643 × 105 N/C .

The direction must be toward the rod, since the charge distribution is negative (a positive test charge would be attracted), so the sign should be positive, according to the convention stated in the problem.

Correct answer: 8.20643 × 105 N/C. Explanation: Let : ℓ = 21 cm = 0.21 m , Q = −25.7 µC = −2.57 × 10−5 C , r = 54.0821 cm = 0.540821 m , and ke = 8.98755 × 109 N · m2 /C2 . For a rod of length ℓ and linear charge density (charge per unit length) λ, the field at a distance d from the end of the rod along the axis is  Z d+ℓ λ −λ d+ℓ E = ke dx = ke x2 x  d

1

002 10.0 points A 5.1 g piece of Styrofoam carries a net charge of −0.5 µC and floats above the center of a very large horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet? The acceleration due to gravity is 9.8 m/s2 and the permittivity of free space is 8.85419 × 10−12 C2/N/m2.

d

λℓ , = ke d (ℓ + d)

where dq = λ dx. The linear charge density Q so that (if the total charge is Q) is λ = ℓ Q ℓ ke Q ℓ E = ke = . d (ℓ + d) d (ℓ + d) In this problem, we have the following situation (the distance r from the center is given):

Correct answer: −1.77013 µC/m2. Explanation:

Let :

m = 5.1 g = 0.0051 kg , q = −0.5 µC = −5 × 10−7 C , g = 9.8 m/s2 , and ǫ0 = 8.85419 × 10−12 C2 /N/m2 .

σ due to a nonconduct2 ǫ0 ing infinite sheet of charge is the same as that very close to any plane uniform charge distribution, where σ is the surface charge density (charge per unit area) of the plastic sheet. The floating styrofoam must be in equilibrium, so the electric force must cancel the The field E =

The distance d is 0.21 m ℓ d = r − = 0.540821 m − 2 2 = 0.435821 m ,

force of gravity and Fg = q E σ mg = q 2 ǫ0 mg σ = 2 ǫ0 q

bausinger (nlb934) – HW 2 – shih – (55050)  Ki  qE = m m ∆x Ki Ki = E= qp ∆x q ∆x

= 2 (8.85419 × 10−12 C2 /N/m2) (0.0051 kg) (9.8 m/s2 ) 106 µC · × −5 × 10−7 C 1C = −1.77013 µC/m2 . 003 10.0 points Each proton in a particle beam has a kinetic energy of 4.35 × 10−15 J. What electric field strength will stop these protons in a distance of 1.99 m? The mass of a proton is 1.6726 × 10−27 kg and the fundamental charge is 1.602 × 10−19 C. Correct answer: 13645 N/C. Explanation: Let : K = 13645 N/C , ∆x = 1.99 m , and qp = 1.602 × 10−19 C . The kinetic energy is due to the initial velocity: 1 K = m v2i 2 2 = 2K . vi m From kinematics, vf2 = vi2 + 2 a∆x = 0 2 Ki + 2 a ∆x = 0 m −Ki a= . m ∆x The electric field supplies the net deceleration, so Fe = Fnet q E = m |a|

=

2

4.35 × 10−15 m (1.602 × 10−19 C)(1.99 m)

= 13645 N/C , directed opposite the proton’s velocity. 004 10.0 points A 70 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.66 × 105 N · m2 /C. What is the electric field strength? Correct answer: 1.47072 × 106 N/C. Explanation: Let : r = 35 cm = 0.35 m and Φ = 5.66 × 105 N · m2 /C . I ~ · dA ~ . The position of By Gauss’ law, Φ = E

maximum electric flux will be that position in which the plane of the loop is perpendicular ~ · dA ~ = E dA. to the electric field; i.e., when E Since the field is constant,

Φ = E A = E π r2 5.66 × 105 N · m2 /C Φ = E= π r2 π (0.35 m)2 = 1.47072 × 106 N/C . 005 (part 1 of 2) 10.0 points A cylindrical shell of radius 8.5 cm and length 264 cm has its charge density uniformly distributed on its surface. The electric field intensity at a point 28.9 cm radially outward from its axis (measured from the midpoint of the shell ) is 55200 N/C. What is the net charge on the shell? The Coulomb constant is 8.99 × 109 N · m2 /C2.

bausinger (nlb934) – HW 2 – shih – (55050) Correct answer: 2.34235 × 10−6 C.

3 Q2

Q′′2

Explanation:

Q2′

Q1 Let : a = 0.085 m , ℓ = 2.64 m , E = 55200 N/C , r = 0.289 m , and

b

a

P

c

k = 8.99 × 109 N · m2/C2 . Applying Gauss’ law Q ǫ0 Q 2πrℓE = ǫ0

I

E · dA =

Find the charge Q′′2 . 1. Q′′2 = Q1 + Q2 correct 2. Q′′2 =

Q1 − Q2 2

3. Q′′2 = Q2 − Q1 Q 2Q 2kQ E= = = rℓ 2 π ǫ0 r ℓ 4 π ǫ0 r ℓ Erℓ Q= 2k (55200 N/C) (0.289 m) (2.64 m) = 2 (8.99 × 109 N · m2 /C2) = 2.34235 × 10−6 C .

(Q1 + Q2)2 Q1 − Q2 Q + Q2 5. Q′′2 = 1 2 Q − Q1 6. Q′′2 = 2 2 4. Q′′2 =

7. Q′′2 = Q1 − Q2 8. Q′′2 = 2 (Q2 − Q1)

006 (part 2 of 2) 10.0 points What is the electric field at a point 4.72 cm from the axis?

9. Q′′2 = 2 (Q1 − Q2) 10. Q′′2 = 2 (Q1 + Q2)

Correct answer: 0 N/C. Explanation: E = 0 inside the shell. 007 (part 1 of 2) 10.0 points Consider a solid conducting sphere with a radius a and charge Q1 on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius b > a, outer radius c and a net charge Q2 on the shell. Denote the charge on the inner surface of the shell by Q2′ and that on the outer surface of the shell by Q′′2 .

Explanation: Sketch a concentric Gaussian surface S (dashed line) within the shell.

bausinger (nlb934) – HW 2 – shih – (55050) Q2 Q1

4 ke (Q1 − Q2) (a + b)2 2 ke Q2 10. EP = (a + b)2 Explanation: Choose as your Gaussian surface the spherical surface S concentric with the centers of the spheres, which passes through P : 9. EP =

Q′′2

r

4

Q2′ Q1

Q2 Since the electrostatic field in a conducting medium is zero, according to Gauss’s Law,

r

Q2′′

Q1 P

Q1 + Q′2 =0 ǫ0 Q′2 = −Q1

Q′2 Q1

ΦS =

The net charge on the shell is Q2 = Q′2 + Q2′′ Q′′2 = Q2 − Q′2 = Q2 + Q1 . 008 (part 2 of 2) 10.0 points Find the magnitude of  the electric field at ~ point P kEP k ≡ EP , where the distance a+b from P to the center is r = . 2 1. EP = 0 2. EP = 3. EP = 4. EP = 5. EP = 6. EP = 7. EP = 8. EP =

4 ke (Q1 + Q2 ) (a + b)2 4 ke Q2 (a + b)2 4 ke Q1 correct (a + b)2 2 ke (Q1 + Q2 ) (a + b)2 2 ke (Q1 − Q2 ) (a + b)2 2 ke Q1 a (a + b)3 2 ke Q1 (a + b)2

Q1 ǫ0 Q1 ke Q1 4 ke Q1 EP = = = . 2 2 4 π ǫ0 r r (a + b)2

4 π r2 EP =

009 (part 1 of 5) 10.0 points Consider a cylindrical charge distribution extending from r = 0 to r = 9.5 cm of charge r density ρ = , where a0 = 28 cm · m3/µC. a0 This cylindrical charge distribution is surrounded by a dielectric shell of dielectric constant 6.31 whose inner radius is 16.1 cm and outer radius is 25 cm.

L

9.5 cm

16.1 cm

25 cm

What is the electric field at 2.81 cm? Assume the length L is very long compared to the diameter of the dielectric shell and neglect

bausinger (nlb934) – HW 2 – shih – (55050)

5

edge effects. The permittivity of a vacuum is 8.8542 × 10−12 C2 /N · m2 . Let : Correct answer: 106.166 V/m. Explanation: Let :

R = 9.5 cm , r = 2.81 cm , a0 = 28 cm · m3/µC , and ǫ0 = 8.8542 × 10−12 C2/N · m2

L

2R

2 rin

2 rout

We are between the charge distribution and the dielectric, so the integration in the RHS of Gauss’ law goes up to the boundary R of the charge distribution: Z R 1 2πrE = r′ (2 π r′ ) dr′ ǫ0 a0 0 R3 E= 3 ǫ0 a0 r 1 = 8.8542 × 10−12 C2 /N · m2 (9.5 cm)3 × 3 (28 cm · m3/µC) (14 cm) 1m 1C × × 100 cm 1 × 106 µC = 823.406 V/m .

  dV = d π r2 L = (2 π r) L dr .

Applying Gauss’ law inside the charge distribution, I Z r Q 1 encl ~E d~a = = ρ(r′ ) dV ǫ0 0 ǫ0 Z r 1 r′ · 2 π r′ L dr 2π rLE = ǫ0 a0 0 r2 r3 E= = 3 ǫ0 a0 r 3 ǫ0 a0 1 = 8.8542 × 10−12 C2 /N · m2 (2.81 cm)2 × 3 (28 cm · m3/µC) 1C 1m × × 100 cm 1 × 106 µC = 106.166 V/m . 010 (part 2 of 5) 10.0 points What is the electric field at 14 cm? Correct answer: 823.406 V/m. Explanation:

R = 9.5 cm and r = 14 cm .

011 (part 3 of 5) 10.0 points What is the electric field at 20.9 cm? Correct answer: 87.4111 V/m. Explanation: Let : r = 20.9 cm κ = 6.31 .

and

Now we are inside the dielectric which will 1 decrease the electric field by a factor of : κ E= =

R3 3 ǫ0 a0 κ r 1 C2 /N · m2 (9.5 cm)3 × 3 (28 cm · m3/µC) (6.31) (20.9 cm) 1C 1m × × 100 cm 1 × 106 µC 8.8542 × 10−12

= 87.4111 V/m .

bausinger (nlb934) – HW 2 – shih – (55050) 012 (part 4 of 5) 10.0 points What is the electric field at 33.4 cm? Correct answer: 345.14 V/m. Explanation: Let : r = 33.4 cm . Now we are outside of the dielectric; although there is induced surface charge on the shell, the net charge on it is zero, so we can disregard the shell when applying Gauss’ law: R3 E= 3 ǫ0 a0 r 1 = −12 8.8542 × 10 C2 /N · m2 (9.5 cm)3 × 3 (28 cm · m3/µC) (33.4 cm) 1C 1m × × 100 cm 1 × 106 µC = 345.14 V/m .

Correct answer: −0.00533496 µC/m2. Explanation: The electric field jumps from E(rin) just E(rin) inside the dibelow the dielectric to κ electric at its boundary, so the surface charge density induced at the boundary is   1 σ = −ǫ0 E(rin) 1 − , κ R3 , 3 ǫ0 a0 rin

014 (part 1 of 2) 10.0 points A uniformly charged conducting plate with area A has a total charge Q which is positive. The figure below shows a cross-sectional view of the plane and the electric field lines due to the charge on the plane. The figure is not drawn to scale. +Q E

E

P

Find the magnitude of the field at point P, which is a distance a from the plate. Assume that a is very small when compared to the dimensions of the plate, such that edge effects can be ignored. ~ P k = 4 π ǫ0 a2 Q 1. kE

013 (part 5 of 5) 10.0 points What is the surface charge density on the inner surface of the dielectric?

where E(rin) =

6

so

κ − 1 R3 3 κ a0 rin (9.5 cm)3 =− (28 cm · m3 /µC) (16.1 cm) 6.31 − 1 1 m × 3 (6.31) 100 cm

~ P k = 2 ǫ0 Q A 2. kE ~ P k = ǫ0 Q A 3. kE ~ P k = ǫ0 Q a2 4. kE ~ P k = 4 π ǫ0 a Q 5. kE Q 4 π ǫ0 a2 ~ Pk = Q 7. kE 4 π ǫ0 a ~ Pk = Q 8. kE ǫ0 A ~ P k = Q correct 9. kE 2 ǫ0 A ~ Pk = 6. kE

σ=−

= −0.00533496 µC/m2 .

Explanation: Basic Concepts Gauss’ Law, electrostatic properties of conductors. Solution: Consider the Gaussian surface shown in the figure.

bausinger (nlb934) – HW 2 – shih – (55050) +Q

E

E S

Due to the symmetry of the problem, there is an electric flux only through the right and left surfaces and these two are equal. Total Q charge density is σ = ; each (large) conducA tor surface holds half the total charge, so each large surface has half the total charge density: Q . If the cross section of the surface σL,R = 2A is S, then Gauss’ Law states that ΦTOTAL = 2 E S 1 Q (2S) , = ǫ0 2A I ~ · dA ~ , E

ΦTOTAL ≡ E=

since so

Q . 2 ǫ0 A

015 (part 2 of 2) 10.0 points Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q while plate #2 has a charge −3 Q. +Q −3 Q y x

P

#1

Q ǫ0 A ~ P k = 2 Q correct 2. kE ǫ0 A ~ Pk = 1. kE

~ Pk = 4 Q 3. kE ǫ0 A ~ Pk = 3 Q 4. kE 2 ǫ0 A ~ Pk = 3 Q 5. kE ǫ0 A ~ Pk = Q 6. kE ǫ0 ~ Pk = Q 7. kE 2 ǫ0 A ~ Pk = 5 Q 8. kE ǫ0 A ~ Pk = Q 9. kE 3 ǫ0 A ~ Pk = 0 10. k E Explanation: According to the result of part 1, the electric field generated by plate #1 at P is Q directed along the positive x-axis. E1 = 2 ǫ0 A The field generated by plate #2 has a magni3Q and it is directed along the tude E2 = 2 ǫ0 A positive x-axis also, since the charge is negative. Therefore, the magnitudes add and we get Egap =

3Q 2Q Q + = . 2 ǫ0 A 2 ǫ0 A ǫ0 A

016 10.0 points Two charges, −2 Q and +Q, are located on the x-axis, as shown. Point P , at a distance of 3 D from the origin O, is one of two points on the positive x-axis at which the electric potential is zero.

#2

Using the superposition principle find the magnitude of the electric field at a point P in the gap.

7

D 0

D −2Q

D +Q

x P

How far from the origin O is the other point? The Coulomb constant is 8.98755 × 109 N · m2 /C2 . 1. d =

7 D 4

bausinger (nlb934) – HW 2 – shih – (55050)

What is the absolute value of the change in potential from the origin to point A?

2. None of these 6 D 5 7 4. d = D 5 3 5. d = D 2 5 6. d = D 4 4 7. d = D 3 8 8. d = D 5 9 9. d = D 5 5 10. d = D correct 3 Explanation: Suppose the zero-potential point is at a distance d from O. The electric potentials Q −2 Q and are , respec4 π ǫ0 |D − d| 4 π ǫ0 |2D − d| tively. For zero electric potential 3. d =

Correct answer: 100.92 V. Explanation: Let : x = 29 cm , y = 51 cm , and ~ = 348 V/m . kEk The potential difference from O to A is defined as ∆V = VA − VO = −

I

or

~ E · d~s .

(x, y) A

O

B

I

VA − VO = ( VA − VB ) + ( VB − VO ) ,

d = 3D.

017 10.0 points A uniform electric field of magnitude 348 V/m is directed in the positive x-direction. Suppose a 25 µC charge moves from the origin to point A at the coordinates, (29 cm, 51 cm). y 348 V/m (29 cm, 51 cm) A

~ and d~s are both along the From O to B, E ~ · d~s = E dx. From B to A, E ~ and x-axis, so E ~ d~s are perpendicular, so E · d~s = 0. Z A ~ ~E · d~s VA − VO = − E · d~s − O B Z y Z x 0 dy E dx − =− 0 0 Z x dx = −E ∆x = −E Z

B

O

O

x

Path I:

which gives 5 D 3

A

~ = (348 V/m) ˆı . We need We know that E to choose a path to integrate along. Because the electric force is conservative, it doesn’t matter which path we take; they all give the same answer. There are two choices of path for which the math is simple (see the figure below.) y E

II

1 −2 , = |2D − d| |D − d|

Z

O

Q −2 Q = 0, + 4 π ǫ0 |D − d| 4 π ǫ0 |2 D − d|

d=

8

x

= −(348 V/m) (0.29 m) = −100.92 V .

bausinger (nlb934) – HW 2 – shih – (55050) The absolute value is |∆V | = 100.92 V . ~ · d~s = E cos θ ds . Path II: In this case, E x where cos θ = ⇒ x = l cos θ . l Z l VA − VO = −E cos θ ds O

= −E l cos θ = −E x . which is the same as the result for the other path.

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