Download MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 11

MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 11: ELECTRIC FIELDS AND FORCES
WORKSHEET SOLUTIONS
1. At what point is the electrical field associated with a uniformly-charged, hollow, metallic sphere
greatest?
(A) center of the sphere
(B) at the sphere’s inner surface
(C) at infinity
(D) at the sphere’s outer surface
(E) within the metal shell
1A. (D) at the sphere’s outer surface
2. Two point charges are placed along a horizontal axis with the following values and positions: +3 C
at x = 0 cm and -7 C at x = 20 cm ( = “micro” = 10-6). What is the magnitude of the electric field at
the point midway between the two charges (at x = 10 cm)?
2A.
(1) E = E1 + E2
(2) E1 = kq1 / r2
(3) E1 = (9.0 x 109 N.m2/C2)(3 x 10-6 C) / (0.10 m)2
(4) E1 = 2.7 x 106 N/C
(5) E2 = kq2 / r2
(6) E2 = (9.0 x 109 N.m2/C2)(7 x 10-6 C) / (0.10 m)2
(7) E2 = 6.3 x 106 N/C
(8) E = 2.7 x 106 N/C + 6.3 x 106 N/C
(9) E = 9.0 x 106 N/C
3. Initially, a net charge of – 5.0 C is placed on a 5.0 cm radius metallic hollow sphere. Next, a +10
C charge is carefully inserted at the center through a hole in the latter’s surface. What electric field is
present at a point 10 cm from the center of the sphere?
3A.
(1) E = k(QEFF) / r2
(2) E = (9.0 x 109 N.m2/C2)(5 x 10-6 C) / (0.1 m)2
(3) E = 4.5 x 106 N/C
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4. Two point charges are separated by 5 cm and have charges of + 3.0 C and – 4.0 C, respectively.
What is the electric field at a point midway between the two charges?
4A.
(1) E = E1 + E2
(2) E1 = kq1 / r2
(3) E1 = (9.0 x 109 N.m2/C2)(3 x 10-6 C) / (0.025m)2
(4) E1 = 4.32 x 107 N/C
(5) E2 = kq2 / r2
(6) E2 = (9.0 x 109 N.m2/C2)(4 x 10-6 C) / (0.025m)2
(7) E2 = 5.76 x 107 N/C
(8) E = 4.32 x 107 N/C + 5.76 x 107 N/C
(9) E = 1.01 x 108 N/C
5. A Van De Graaff generator has a spherical dome of radius 42 cm. Operating in dry air, where
“atmospheric breakdown” is Emax = 3.00 x 106 N/C what is the maximum charge that can be held on the
dome?
5A.
(1) E = kq / r2
(2) q = Er2 / k
(3) q = (3.00 x 106 N/C)(0.42 m)2 / (9.0 x 109 N.m2/C2)
(4) q = 5.88 x 10-5 C
(5) q = 58.8 C
6. A glass rod rubbed with a silk cloth acquires a charge of + 7.40 x 10-9 C. How many electrons were
transferred from the glass to the silk?
6A.
(7.40 x 10-9 C)(1 electron / 1.6 x 10-19 C)
= 4.63 x 1010 electrons
7. A positive point charge is brought near a neutral conducting sphere. The sphere is isolated from
ground. Sketch the induced charge on the sphere
7A. The positive charge attracts negative charge to the left side of the sphere leaving positive charges
on the right side:
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8. A positive point charge is brought near a grounded conductor. Sketch the induced charge on the
conductor’s surface:
8A.
The positive charge attracts electrons from ground to the sphere’s surface – the closer the charge is to
the sphere the more negative it becomes:
9. What happens to the charge on the sphere when the positive charge is removed?
9A.
It runs back to ground through the wire joining it to the earth. A charged conductor is made neutral by
grounding it.
10. Three point charges are on the x-axis as shown below. Compute the net force on the 30 micro
(30x) coulomb charge.
10A.
(1) Force from the left: F = kq1q2 / r2
(2) F = (9.0 x 109 N.m2/C2)(20 x 10-6 C)
(30 x 10-6 C) / (0.4 m)2
(3) F = 33.75 N (to the right)
(4) Force from the right: F = kq1q2 / r2
(5) F = (9.0 x 109 N.m2/C2)(25 x 10-6 C)(30 x 10-6 C) / (0.6 m)2
(6) F = - 18.75 N (to the left)
(7) FNET = 33.75 N – 18.75 N
(8) FNET = 15 N to the right
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CHAPTER 12: ELECTRIC ENERGY
WORKSHEET SOLUTIONS
1. At which location will the magnitude of the electric field between the two parallel plates of a charged
capacitor be the greatest?
(A) near the positive plate
(B) near the negative plate
(C) midway between the two plates
(D) electric field is constant
(E) midway between midpoint and positive plate
1A. (D) electric field is constant
2. When moving an electrical charge from one point to another in the presence of an electrical field,
which quantity depends on the size of the charge that is moved?
(A) electric field
(B) work done
(C) potential difference
(D) distance moved
(E) magnetic field
2A. (B) work done
3. Two capacitors with charges of 1.00 and 0.50 F, respectively, are connected in parallel. The system
is connected to a 100 V battery. What charge accumulates on the 1.00 F capacitor?
3A.
(1)
(2)
(3)
(4)
C=Q/V
Q = CV
Q = (1.00 x 10-6 F)(100 V)
Q = 100 C
4. A 120 V battery is connected to a 3.50 F, parallel plate, air-filled capacitor. Now the battery is
disconnected, with care taken not to discharge the plates. Some Pyrex glass is next inserted between the
two plates, completely filling up the space. What is the final potential difference between the plates?
(dielectric constant  = 5.60)
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VAN NUYS HIGH SCHOOL
4A.
(1) Determine initial charge: C = Q / V
(2) Q = CV
(3) Q = (3.5 x 10-6 F)(120 V)
(4) Q = 4.2 x 10-4 C
(5) Determine new capacitance: C’ = o(A/d) = C
(6) C’ = (5.60)(3.5 x 10-6 F)
(7) C’ = 1.96 x 10-5 F
(8) Determine new voltage: C’ = Q / V’
(9) V’ = Q / C’
(10) V’ = (4.2 x 10-4 C) / (1.96 x 10-5 F)
(11) V’ = 21.4 V
5. A parallel plate capacitor has dimensions 3.5 cm x 2.5 cm. The plates are separated by a 1.0 mm
thickness of paper (dielectric constant  = 3.7). What is the charge that can be stored on this capacitor,
when connected to a 12 volt battery?
5A.
(1) C = o(A / d)
(2) A = (2.5 x 10-2 m)(3.5 x 10-2 m)
(3) A = 8.75 x 10-4 m2
(4) C = (3.70)(8.85 x 10-12)(8.75 x 10-4 m2 / 1.00 x 10-3 m)
(5) C = 2.87 x 10-11 F
(6) C = Q / V
(7) Q = CV
(8) Q = (2.87 x 10-11 F)(12 V)
(9) Q = 3.44 x 10-10 C
6. Very large capacitors have been considered as a means for storing electrical energy. If we
constructed a very large parallel plate capacitor of plate area 3.25 m2 using pyrex ( = 5.60) of thickness
1.75 mm as a dielectric, how much electrical energy would it store at a plate voltage of 7000 V?
6A.
(1) Determine the capacitance: C = o(A / d)
(2) C = (5.60) (8.85 x 10-12)(3.25 m2 / 1.75 x 10-3 m)
(3) C = 9.2 x 10-8 F
(4) Determine electrical energy: U = ½ CV2
(5) U = ½ (9.2 x 10-8 F)(7000 V)2
(6) U = 2.25 J
7. There is a hollow, conducting, uncharged sphere with a charge +Q inside the sphere. Consider the
potential and the electrical field at a point P1 inside the metal of the sphere. At this point:
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7A.
(A) only the electrical field is zero.
8. What is the equivalent capacitance of the combination shown?
8A.
(1) Left side: CL = 10 F + 20 F
(2) CL = 30 F
(3) Right side: CR = 30 F + 30 F
(4) CR = 60 F
(5) Both sides: 1 / Ceq = (1 / CL) + (1 / CR)
(6) 1 / Ceq = 1/30 F + 1/60 F
(7) 1 / Ceq = 3/60 F = 1/20 F
(8) Ceq = 20 F
9. Inserting a dielectric material between two charged parallel conducting plates, originally separated by
air and disconnected from a battery, will produce what effect on the capacitor?
9A.
(C) increase capacitance
10. A uniform electric field, with a magnitude of 5 x 102 N/C, is directed parallel to the positive x-axis.
If the potential at x = 5.00 m is 2500 V, what is the potential at x = 2.00 m?
10A.
1000 V. Since the electric field is constant throughout, V is proportional to d. Since x at 5.00 m is equal
to 2500 V, this means 500 V per meter.
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11. A parallel plate capacitor is charged by a battery. The plate area is 15 cm x 250 cm, the gap
between plates is 0.35 mm, and the battery voltage is 30 V.
11a. Calculate the capacitance in nanofarads assuming vacuum between the plates.
A.
(1) C = o(A/d)
(2) A = (L)(w)
(3) A = (0.15 m)(2.5 m)
(4) A = 0.375 m2
(5) C = (8.85 x 10-12)[(0.375 m2) / (3.5 x 10-4 m)
(6) C = 9.48 x 10-9 F = 9.48 nF
11b. Calculate the charge Q on the plates in nanocoulombs.
A.
(1) C = Q / V
(2) Q = CV
(3) Q = (9.48 x 10-9 F)(30 V)
(4) Q = 2.84 x 10-7 C
(5) Q = 284 nC
11c. Evaluate the stored energy in microjoules.
A.
(1) U = ½ CV2
(2) U = ½ (9.48 x 10-9 F)(30 V)2
(3) U = 4.27 J
12. Now fill the region between the plates with material of dielectric constant 3.0 while maintaining 30
V across the plates by keeping them connected to the battery.
12a. Calculate the capacitance in nanofarads.
A.
(1) C = o(A / d)
(2) A = (L)(w)
(3) A = (0.15 m)(2.5 m)
(4) A = 0.375 m2
(5) C = (3)(8.85 x 10-12)[(0.375 m2) / (3.5 x 10-4 m)]
(6) C = 2.84 x 10-8 F = 28.4 nF
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12b. Calculate the charge Q on the plates in nanocoulombs.
A.
(1) C = Q / V
(2) Q = CV
(3) Q = (28.4 x 10-9 F)(30 V)
(4) Q = 8.52 x 10-7 C
(5) Q = 852 nC
12c. Evaluate the stored energy in microjoules.
A.
(1) Uc = ½ CV2
(2) Uc = ½ (28.4 x 10-9 F)(30 V)2
(3) Uc = 1.29 x 10-5 J
(4) Uc = 12.9 J
13. Two capacitors are connected in parallel.
13a. What is the voltage and charge for the 2 microfarad capacitor?
A.
(1) Voltage remains constant at 10 V.
(2) C = Q / V
(3) Q = CV
(4) Q = (2 F)(10 V)
(5) Q = 20 C
13b. What is the voltage and charge for the 4 microfarad capacitor?
(1) Voltage is constant at 10 V.
(2) C = Q / V
(3) Q = CV
(4) Q = (4 F)(10 V)
(5) Q = 40 C
13c. What is the total charge stored?
(1) Q = Q1 + Q2
(2) Q = 20 C + 40 C
(3) Q = 60 C
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13d. Determine the equivalent capacitance of the combination in microfarads.
(1) Ceq = C1 + C2
(2) Ceq = 2 F + 4 F
(3) Ceq = 6 F
14. Two capacitors are connected in series.
14a. What is the equivalent capacitance in microfarads?
A.
(1) 1 / Ceq = 1 / C1 + 1 / C2
(2) 1 / Ceq = 1 / (3 F) + 1 / (6 F)
(3) 1 / Ceq = 2 / (6 F) + 1 / (6 F)
(4) 1 / Ceq = 3 / (6 F)
(5) 1 / Ceq = 1 / (2 F)
(6) Ceq = 2 F
14b. Determine the charge on the 3 microfarad capacitor in microcoulombs.
A.
(1) Ceq = Q1 / V
(2) Q1 = (Ceq)(V)
(3) Q1 = (2 F)(10 V)
(4) Q1 = 20 C
14c. Calculate the voltage across the 3 microfarad capacitor.
A.
(1) C1 = Q1 / V1
(2) V1C1 = Q1
(3) V1 = Q1 / C1
(4) V1 = 20 C / 3 F
(5) V1 = 6.7 V
14d. Calculate the voltage across the 6 microfarad capacitor.
A.
(1) C2 = Q2 / V2
(2) V2C2 = Q2
(3) V2 = Q2 / C2
(4) V2 = 20 C / 6 F
(5) V2 = 3.3 V
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CHAPTERS 11 - 12: ELECTRIC FORCES and ENERGY
QUIZ SOLUTIONS
1. Two point charges are placed along a horizontal axis with the following values and positions:
+4 C at x = 0 cm and -9 C at x = 20 cm. What is the magnitude of the electric field at the point
midway between the two charges (at x = 10 cm)?
1A.
(1) E = E1 + E2
(2) E1 = kq1 / r2
(3) E1 = (9.0 x 109 N.m2/C2)(4 x 10-6 C) / (0.10 m)2
(4) E1 = 3.60 x 106 N/C
(5) E2 = kq2 / r2
(6) E2 = (9.0 x 109 N.m2/C2)(9 x 10-6 C) / (0.10 m)2
(7) E2 = 8.10 x 106 N/C
(8) E = 3.60 x 106 N/C + 8.10 x 106 N/C
(9) E = 1.17 x 107 N/C
2. Initially, a net charge of – 11.0 C is placed on a 5.0 cm radius metallic hollow sphere. Next, a +10
C charge is carefully inserted at the center through a hole in the latter’s surface. What electric field is
present at a point 20 cm from the center of the sphere?
2A.
(1) E = k(QEFF) / r2
(2) QEFF = 10 C – 11 C
(3) QEFF = - 1 C
(4) E = (9.00 x 109 N.m2/C2)(1 x 10-6 C) / (0.2 m)2
(5) E = 2.25 x 105 N/C
3. A Van De Graaff generator has a spherical dome of radius 50 cm. Operating in dry air, where
“atmospheric breakdown” is Emax = 3.00 x 106 N/C what is the maximum charge that can be held on the
dome?
3A.
(1) E = kq / r2
(2) q = Er2 / k
(3) q = (3.00 x 106 N/C)(0.50 m)2 / (9.0 x 109 N.m2/C2)
(4) q = 8.34 x 10-5 C
4. A glass rod rubbed with a silk cloth acquires a charge of + 8.49 x 10-8 C. How many electrons were
transferred from the glass to the silk?
4A. (8.49 x 10-8 C)(1 electron / 1.6 x 10-19 C) = 5.31 x 1011 electrons
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5. A positive point charge is brought near a grounded conductor. Sketch the induced charge on the
conductor’s surface:
5A. The positive charge attracts electrons from ground to the sphere’s surface – the closer the charge is
to the sphere the more negative it becomes:
6. What happens to the charge on the sphere when the positive charge is removed?
6A. It runs back to ground through the wire joining it to the earth. A charged conductor is made
neutral by grounding it.
7. Three point charges are on an x-axis. Charge Q1 on the left is +20 C. Charge Q3 on the right is -30
C. They are 1.2 meters apart. Charge Q2 is midway between them. Calculate the net electric force on
Q2 if it has a +15 C charge.
7A.
(1) Force from left: F = kQ1Q2/r2
(2) (9.0 x 109 N.m2/C2)(20 x 10-6 C)(15 x 10-6 C) / (0.6 m)2
(3) F = 7.5 N (to the right)
(4) Force from right: F = kQ1Q2/r2
(5) (9.0 x 109 N.m2/C2)(30 x 10-6 C)(15 x 10-6 C) / (0.6 m)2
(6) F = 11.25 N (to the right)
(7) F = 18.75 N (to the right)
8. A parallel plate capacitor is charged by a battery. The plate area is 15 cm x 315 cm, the
gap between plates is 1.70 mm, and the battery voltage is 30 V.
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MR. SURRETTE
8a.
A.
(1)
(2)
(3)
(4)
(5)
(6)
8b.
A.
(1)
(2)
(3)
VAN NUYS HIGH SCHOOL
Calculate the capacitance in nanofarads assuming vacuum between the plates.
C = o(A/d)
A = (L)(w)
A = (0.15 m)(3.15 m)
A = 0.47 m2
C = (8.85 x 10-12)[(0.47 m2) / (1.70 x 10-3 m)]
C = 2.46 x 10-9 F = 2.5 nF
Evaluate the stored energy in microjoules.
U = ½ CV2
U = ½ (2.46 x 10-9 F)(30 V)2
U = 1.11 x 10-6 J = 1.1 J
Now fill the region between the plates with material of dielectric constant 4.7 while maintaining 30 V
across the plates by keeping them connected to the battery.
8c. Calculate the capacitance in nanofarads.
A.
(1) C’ = o(A/d)
(2) C’ = C
(3) C’ = (4.7)(2.46 nF)
(4) C’ = 11.6 nF
8d.
A.
(1)
(2)
(3)
(4)
Calculate the charge Q on the plates in nanocoulombs.
C=Q/V
Q = CV
Q = (11.6 x 10-9 F)(30 V)
Q = 3.48 x 10-7 C = 348 nC
9. Two capacitors are connected in series. What is the equivalent capacitance in microfarads?
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9A.
(1) 1 / Ceq = 1 / C1 + 1 / C2
(2) 1 / Ceq = 1 / (3 F) + 1 / (6 F)
(3) 1 / Ceq = 2 / (6 F) + 1 / (6 F)
(4) 1 / Ceq = 3 / (6 F)
(5) 1 / Ceq = 1 / (2 F)
(6) Ceq = 2 F
10. A 175 V battery is connected to a 1.15 F, parallel plate, air-filled capacitor. Now the battery is
disconnected, with care taken not to discharge the plates. Some Pyrex glass is next inserted between the
two plates, completely filling up the space. What is the final potential difference between the plates?
(dielectric constant  = 5.60)
10A.
(1) Determine initial charge: C = Q / V
(2) Q = CV
(3) Q = (1.15 x 10-6 F)(175 V)
(4) Q = 2.01 x 10-4 C
(5) Determine new capacitance: C’ = o(A/d) = C
(6) C’ = (5.60)(1.15 x 10-6 F)
(7) C’ = 6.44 x 10-6 F
(8) Determine new voltage: C’ = Q / V’
(9) V’ = Q / C’
(10) V’ = (2.01 x 10-4 C) / 6.44 x 10-6 F)
(11) V’ = 31.3 V
11. A parallel plate capacitor has dimensions 4.52 cm x 3.15 cm. The plates are separated by a 1.00
mm thickness of paper (dielectric constant  = 3.70). What is the charge that can be stored on this
capacitor, when connected to a 12.0 volt battery?
11A.
(1) C = o(A / d)
(2) A = (4.52 x 10-2 m)(3.15 x 10-2 m)
(3) A = 1.42 x 10-3 m2
(4) C = (3.70)(8.85 x 10-12)(1.42 x 10-3 m2 / 1.00 x 10-3 m)
(5) C = 4.65 x 10-11 F
(6) C = Q / V
(7) Q = CV
(8) Q = (4.65 x 10-11 F)(12.0 V)
(9) Q = 5.6 x 10-10 C
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12. Very large capacitors have been considered as a means for storing electrical energy. If we
constructed a very large parallel plate capacitor of plate area 5.12 m2 using pyrex ( = 5.60) of thickness
1.14 mm as a dielectric, how much electrical energy would it store at a plate voltage of 16,000 V?
12A.
(1) Determine the capacitance: C = o(A / d)
(2) C = (5.60)(8.85 x 10-12 C2/N.m2)(5.12 m2 / 1.14 x 10-3 m)
(3) C = 2.23 x 10-7 F
(4) Determine the electrical energy: U = ½CV2
(5) U = ½(2.23 x 10-7 F)(16,000 V)2
(6) U = 28.5 J
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