CAPACITANCE AND DIELECTRICS – PARALLEL PLATE
CAPACITOR
1 INTRODUCTION
This document presents a series of numerical explorations related to capacitance and
dielectrics. Each exploration is accompanied by a detailed analysis and solution.
2 EXPLORATION 1: PARALLEL PLATE CAPACITOR
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
1. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
2. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
3 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
1. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
2. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
4 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
1. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
2. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
5 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
1. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
2. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
6 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
1. Use the formula: 𝑄 = 𝐶𝑉
2. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
7 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
1. For capacitors in parallel, add the individual capacitances
2. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
8 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
1. Treat this as two capacitors in parallel: one with air, one with dielectric
2. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
9 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
1. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
2. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
3. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
4. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
10 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
5. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
6. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
11 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
7. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
8. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
12 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
9. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
10. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
13 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
11. Use the formula: 𝑄 = 𝐶𝑉
12. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
14 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
13. For capacitors in parallel, add the individual capacitances
14. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
15 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
15. Treat this as two capacitors in parallel: one with air, one with dielectric
16. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
16 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
17. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
18. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
19. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
20. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
17 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
21. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
22. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
18 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
23. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
24. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
19 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
25. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
26. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
20 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
27. Use the formula: 𝑄 = 𝐶𝑉
28. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
21 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
29. For capacitors in parallel, add the individual capacitances
30. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
22 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
31. Treat this as two capacitors in parallel: one with air, one with dielectric
32. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
23 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
33. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
34. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
35. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
36. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
24 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
37. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
38. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
25 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
39. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
40. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
26 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
41. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
42. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
27 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
43. Use the formula: 𝑄 = 𝐶𝑉
44. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
28 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
45. For capacitors in parallel, add the individual capacitances
46. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
29 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
47. Treat this as two capacitors in parallel: one with air, one with dielectric
48. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
30 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
49. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
51. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
52. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
31 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
53. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
54. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
32 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
55. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
56. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
33 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
57. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
58. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
34 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
59. Use the formula: 𝑄 = 𝐶𝑉
60. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
35 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
61. For capacitors in parallel, add the individual capacitances
62. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
36 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
63. Treat this as two capacitors in parallel: one with air, one with dielectric
64. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
37 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
65. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
66. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
67. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
68. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
38 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
69. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
70. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
39 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
71. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
72. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
40 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
73. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
74. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
41 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
75. Use the formula: 𝑄 = 𝐶𝑉
76. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
42 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
77. For capacitors in parallel, add the individual capacitances
78. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
43 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
79. Treat this as two capacitors in parallel: one with air, one with dielectric
80. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
44 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
81. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
82. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
83. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
84. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
45 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
85. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
86. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
46 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
87. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
88. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
47 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
89. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
90. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
48 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
91. Use the formula: 𝑄 = 𝐶𝑉
92. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
49 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
93. For capacitors in parallel, add the individual capacitances
94. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
50 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
95. Treat this as two capacitors in parallel: one with air, one with dielectric
96. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
51 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
97. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
98. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
99. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
100. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
52 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
101. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
102. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
53 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
103. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
104. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
54 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
105. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
106. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
55 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
107. Use the formula: 𝑄 = 𝐶𝑉
108. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
56 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
109. For capacitors in parallel, add the individual capacitances
110. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
57 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
111. Treat this as two capacitors in parallel: one with air, one with dielectric
112. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
58 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
113. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
114. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
115. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
116. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
59 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
117. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
118. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
60 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
119. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
120. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
61 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
121. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
122. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
62 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
123. Use the formula: 𝑄 = 𝐶𝑉
124. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
63 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
125. For capacitors in parallel, add the individual capacitances
126. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
64 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
127. Treat this as two capacitors in parallel: one with air, one with dielectric
128. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
65 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
129. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
130. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
131. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
132. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
66 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
133. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
134. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
67 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
135. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
136. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
68 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
137. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
138. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
69 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
139. Use the formula: 𝑄 = 𝐶𝑉
140. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
70 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
141. For capacitors in parallel, add the individual capacitances
142. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
71 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
143. Treat this as two capacitors in parallel: one with air, one with dielectric
144. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
72 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
145. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
146. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
147. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
148. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
73 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
149. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
150. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
74 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
151. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
152. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
75 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
153. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
154. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
76 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
155. Use the formula: 𝑄 = 𝐶𝑉
156. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
77 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
157. For capacitors in parallel, add the individual capacitances
158. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
78 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
159. Treat this as two capacitors in parallel: one with air, one with dielectric
160. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
79 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
161. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
162. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
163. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
164. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
80 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
165. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
166. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
81 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
167. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
168. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
82 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
169. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
170. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
83 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
171. Use the formula: 𝑄 = 𝐶𝑉
172. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
84 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
173. For capacitors in parallel, add the individual capacitances
174. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
85 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
175. Treat this as two capacitors in parallel: one with air, one with dielectric
176. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
86 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
177. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
178. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
179. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
180. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
87 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
181. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
182. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
88 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
183. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
184. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
89 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
185. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
186. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
90 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
187. Use the formula: 𝑄 = 𝐶𝑉
188. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
91 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
189. For capacitors in parallel, add the individual capacitances
190. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
92 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
191. Treat this as two capacitors in parallel: one with air, one with dielectric
192. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
93 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
193. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
194. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
195. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
196. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
94 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
197. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
198. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
95 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
199. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
200. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
96 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
201. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
202. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
97 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
203. Use the formula: 𝑄 = 𝐶𝑉
204. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
98 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
205. For capacitors in parallel, add the individual capacitances
206. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
99 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
207. Treat this as two capacitors in parallel: one with air, one with dielectric
208. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
100 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
209. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
210. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
211. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
212. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
101 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
213. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
214. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
102 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
215. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
216. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
103 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
217. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
218. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
104 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
219. Use the formula: 𝑄 = 𝐶𝑉
220. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
105 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
221. For capacitors in parallel, add the individual capacitances
222. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
106 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
223. Treat this as two capacitors in parallel: one with air, one with dielectric
224. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
107 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
225. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
226. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
A parallel plate capacitor has plates of area 0.02 m2 separated by 0.5 mm. Calculate its
capacitance in air.
Analysis:
227. Use the formula for parallel plate capacitor: 𝐶 = 𝜖0𝐴
𝑑
228. 𝜖0= 8.85 ×10−12 F/m
Solution:
𝐶 = 𝜖0𝐴
𝑑
=(8.85 ×10−12)(0.02)
0.5 × 10−3
=354 ×10−12 F=354 pF
108 EXPLORATION 2: ENERGY STORED
A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in the
capacitor?
Analysis:
229. Use the formula for energy stored in a capacitor: 𝑈 = 1
2𝐶𝑉2
230. Pay attention to units and convert if necessary
Solution:
𝑈 = 1
2𝐶𝑉2
=1
2(2 × 10−6)(100)2
= 1 × 10−2 J=10 mJ
109 EXPLORATION 3: CAPACITORS IN SERIES
Three capacitors of 2 μF, 4 μF, and 8 μF are connected in series. What is the equivalent
capacitance?
Analysis:
231. Use the formula for capacitors in series: 1
𝐶𝑒𝑞 =1
𝐶1+1
𝐶2+1
𝐶3
232. Solve for 𝐶𝑒𝑞
Solution:
1
𝐶𝑒𝑞 =1
2+1
4+1
8
=4
8+2
8+1
8=7
8
𝐶𝑒𝑞 =8
7 μF ≈ 1.14 μF
110 EXPLORATION 4: DIELECTRIC EFFECT
A parallel plate capacitor has a capacitance of 5 pF in air. When a dielectric is inserted between
the plates, the capacitance increases to 20 pF. What is the dielectric constant of the material?
Analysis:
233. The ratio of capacitances is equal to the dielectric constant: 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡 = 𝜅
234. Solve for 𝜅
Solution:
𝜅 = 𝐶𝑤𝑖𝑡ℎ
𝐶𝑤𝑖𝑡ℎ𝑜𝑢𝑡
=20 ×10−12
5 × 10−12 = 4
111 EXPLORATION 5: CHARGE ON CAPACITOR
A 3 μF capacitor is connected to a 12 V battery. What is the charge stored on each plate of the
capacitor?
Analysis:
235. Use the formula: 𝑄 = 𝐶𝑉
236. Ensure consistent units
Solution:
𝑄 = 𝐶𝑉
=(3 × 10−6)(12)
=36 ×10−6 C=36 μC
112 EXPLORATION 6: CAPACITORS IN PARALLEL
Calculate the equivalent capacitance of 4 μF, 6 μF, and 12 μF capacitors connected in parallel.
Analysis:
237. For capacitors in parallel, add the individual capacitances
238. 𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
Solution:
𝐶𝑒𝑞 = 𝐶1+ 𝐶2+ 𝐶3
= 4 + 6 + 12
=22 μF
113 EXPLORATION 7: CAPACITOR WITH DIELECTRIC PARTIALLY INSERTED
A parallel plate capacitor with plate area 0.01 m2 and separation 2 mm has a dielectric of
constant 3 inserted halfway between the plates. Calculate the capacitance.
Analysis:
239. Treat this as two capacitors in parallel: one with air, one with dielectric
240. Calculate each capacitance and add them
Solution:
𝐶𝑎𝑖𝑟 = 𝜖0
𝐴/2
𝑑=(8.85 ×10−12)0.005
0.002 =22.125 ×10−12 F
𝐶𝑑𝑖𝑒𝑙 = 𝜅𝜖0
𝐴/2
𝑑= 3(8.85 ×10−12)0.005
0.002 =66.375 × 10−12 F
𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶𝑎𝑖𝑟 + 𝐶𝑑𝑖𝑒𝑙 =88.5 × 10−12 F=88.5 pF
114 EXPLORATION 8: CAPACITOR DISCHARGE
A 10 μF capacitor charged to 200 V is discharged through a 20 kΩ resistor. How long does it
take for the voltage to drop to 50 V?
Analysis:
241. Use the capacitor discharge equation: 𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
242. Solve for 𝑡
Solution:
𝑉 = 𝑉0𝑒−𝑡/𝑅𝐶
50 =200𝑒−𝑡/(20000×10−6)
ln(0.25)= − 𝑡
0.2
𝑡 = −0.2ln(0.25)
= 0.2773 s≈277 ms
