At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Let's address each part of the problem step-by-step.
### Part (a)
Given:
- Area ([tex]\( S \)[/tex]) = 25 cm²
- Separation ([tex]\( d \)[/tex]) = 0.5 mm
- Air dielectric
Solution:
1. Convert the given units to standard SI units:
- [tex]\( S = 25 \times 10^{-4} \)[/tex] m²
- [tex]\( d = 0.5 \times 10^{-3} \)[/tex] m
2. The permittivity of free space ([tex]\( \epsilon_0 \)[/tex]) is [tex]\( 8.854187817 \times 10^{-12} \)[/tex] F/m.
3. Use the formula for capacitance of a parallel-plate capacitor:
[tex]\[ C = \epsilon_0 \frac{S}{d} \][/tex]
4. Substitute the given values into the formula:
[tex]\[ C = 8.854187817 \times 10^{-12} \frac{25 \times 10^{-4}}{0.5 \times 10^{-3}} \][/tex]
[tex]\[ C \approx 4.427 \times 10^{-11} \text{ F} \][/tex]
Therefore, the capacitance is approximately [tex]\( 4.427093908499999 \times 10^{-11} \)[/tex] F.
### Part (b)
Given:
- Area ([tex]\( S \)[/tex]) = 2 m²
- Separation ([tex]\( d \)[/tex]) = 5 mm
- Potential difference ([tex]\( V \)[/tex]) = 10,000 V (10 kV)
- The space between plates is vacuum (i.e., [tex]\( \epsilon_0 \)[/tex])
Solution:
1. Convert the separation [tex]\( d \)[/tex] to meters:
- [tex]\( d = 5 \times 10^{-3} \)[/tex] m
2. (a) Capacitance:
[tex]\[ C = \epsilon_0 \frac{S}{d} \][/tex]
[tex]\[ C = 8.854187817 \times 10^{-12} \frac{2}{5 \times 10^{-3}} \][/tex]
[tex]\[ C \approx 3.542 \times 10^{-9} \text{ F} \][/tex]
The capacitance is approximately [tex]\( 3.5416751267999994 \times 10^{-9} \)[/tex] F.
3. (b) Charge on each plate:
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = 3.542 \times 10^{-9} \times 10,000 \][/tex]
[tex]\[ Q \approx 3.542 \times 10^{-5} \text{ C} \][/tex]
The charge on each plate is approximately [tex]\( 3.541675126799999 \times 10^{-5} \)[/tex] C.
4. (c) Electric field magnitude:
[tex]\[ E = \frac{V}{d} \][/tex]
[tex]\[ E = \frac{10,000}{5 \times 10^{-3}} \][/tex]
[tex]\[ E = 2,000,000 \text{ V/m} \][/tex]
The magnitude of the electric field is approximately [tex]\( 2,000,000.0 \)[/tex] V/m.
### Part (c)
Given:
- Capacitance [tex]\( C_1 = 6 \)[/tex] µF
- Capacitance [tex]\( C_2 = 3 \)[/tex] µF
- Voltage [tex]\( V \)[/tex] = 18 V
Solution:
i) In Series:
1. Equivalent capacitance:
[tex]\[ \frac{1}{C_\text{series}} = \frac{1}{C_1} + \frac{1}{C_2} \][/tex]
[tex]\[ \frac{1}{C_\text{series}} = \frac{1}{6 \times 10^{-6}} + \frac{1}{3 \times 10^{-6}} \][/tex]
[tex]\[ C_\text{series} = 2 \times 10^{-6} \text{ F} \][/tex]
The equivalent capacitance in series is [tex]\( 2 \times 10^{-6} \)[/tex] F (or 2 µF).
2. Charge in series (same for both):
[tex]\[ Q_\text{series} = C_\text{series} \times V \][/tex]
[tex]\[ Q_\text{series} = 2 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_\text{series} = 3.6 \times 10^{-5} \text{ C} \][/tex]
The charge in series is [tex]\( 3.6 \times 10^{-5} \)[/tex] C.
3. Voltage across each capacitor:
[tex]\[ V_{C1(\text{series})} = \frac{Q_\text{series}}{C_1} \][/tex]
[tex]\[ V_{C1(\text{series})} = \frac{3.6 \times 10^{-5}}{6 \times 10^{-6}} \][/tex]
[tex]\[ V_{C1(\text{series})} = 6 \text{ V} \][/tex]
[tex]\[ V_{C2(\text{series})} = \frac{Q_\text{series}}{C_2} \][/tex]
[tex]\[ V_{C2(\text{series})} = \frac{3.6 \times 10^{-5}}{3 \times 10^{-6}} \][/tex]
[tex]\[ V_{C2(\text{series})} = 12 \text{ V} \][/tex]
The voltage across [tex]\( C_1 \)[/tex] is 6 V, and across [tex]\( C_2 \)[/tex] is 12 V.
ii) In Parallel:
1. Equivalent capacitance:
[tex]\[ C_\text{parallel} = C_1 + C_2 \][/tex]
[tex]\[ C_\text{parallel} = 6 \times 10^{-6} + 3 \times 10^{-6} \][/tex]
[tex]\[ C_\text{parallel} = 9 \times 10^{-6} \text{ F} \][/tex]
The equivalent capacitance in parallel is [tex]\( 9 \times 10^{-6} \)[/tex] F (or 9 µF).
2. Charge on each capacitor:
[tex]\[ Q_{C1(\text{parallel})} = C_1 \times V \][/tex]
[tex]\[ Q_{C1(\text{parallel})} = 6 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_{C1(\text{parallel})} = 108 \times 10^{-6} \text{ C} \][/tex]
The charge on [tex]\( C_1 \)[/tex] is [tex]\( 1.08 \times 10^{-4} \)[/tex] C (or 108 µC).
[tex]\[ Q_{C2(\text{parallel})} = C_2 \times V \][/tex]
[tex]\[ Q_{C2(\text{parallel})} = 3 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_{C2(\text{parallel})} = 54 \times 10^{-6} \text{ C} \][/tex]
The charge on [tex]\( C_2 \)[/tex] is [tex]\( 5.4 \times 10^{-5} \)[/tex] C (or 54 µC).
Summary:
- (a) Capacitance: [tex]\(\approx 4.427 \times 10^{-11} \text{ F}\)[/tex]
- (b) Capacitance: [tex]\(\approx 3.542 \times 10^{-9} \text{ F}\)[/tex]
- Charge: [tex]\(\approx 3.542 \times 10^{-5} \text{ C}\)[/tex]
- Electric field: [tex]\(\approx 2,000,000.0 \text{ V/m}\)[/tex]
- (c-i)
- Series Capacitance: [tex]\(2 \times 10^{-6} \text{ F}\)[/tex]
- Charge: [tex]\(3.6 \times 10^{-5} \text{ C}\)[/tex]
- Voltage across [tex]\( C_1 \)[/tex]: 6 V
- Voltage across [tex]\( C_2 \)[/tex]: 12 V
- (c-ii)
- Parallel Capacitance: [tex]\(9 \times 10^{-6} \text{ F}\)[/tex]
- Charge on [tex]\( C_1 \)[/tex]: [tex]\(1.08 \times 10^{-4} \text{ C}\)[/tex]
- Charge on [tex]\( C_2 \)[/tex]: [tex]\(5.4 \times 10^{-5} \text{ C}\)[/tex]
This completes the detailed step-by-step solution for each part of the question.
### Part (a)
Given:
- Area ([tex]\( S \)[/tex]) = 25 cm²
- Separation ([tex]\( d \)[/tex]) = 0.5 mm
- Air dielectric
Solution:
1. Convert the given units to standard SI units:
- [tex]\( S = 25 \times 10^{-4} \)[/tex] m²
- [tex]\( d = 0.5 \times 10^{-3} \)[/tex] m
2. The permittivity of free space ([tex]\( \epsilon_0 \)[/tex]) is [tex]\( 8.854187817 \times 10^{-12} \)[/tex] F/m.
3. Use the formula for capacitance of a parallel-plate capacitor:
[tex]\[ C = \epsilon_0 \frac{S}{d} \][/tex]
4. Substitute the given values into the formula:
[tex]\[ C = 8.854187817 \times 10^{-12} \frac{25 \times 10^{-4}}{0.5 \times 10^{-3}} \][/tex]
[tex]\[ C \approx 4.427 \times 10^{-11} \text{ F} \][/tex]
Therefore, the capacitance is approximately [tex]\( 4.427093908499999 \times 10^{-11} \)[/tex] F.
### Part (b)
Given:
- Area ([tex]\( S \)[/tex]) = 2 m²
- Separation ([tex]\( d \)[/tex]) = 5 mm
- Potential difference ([tex]\( V \)[/tex]) = 10,000 V (10 kV)
- The space between plates is vacuum (i.e., [tex]\( \epsilon_0 \)[/tex])
Solution:
1. Convert the separation [tex]\( d \)[/tex] to meters:
- [tex]\( d = 5 \times 10^{-3} \)[/tex] m
2. (a) Capacitance:
[tex]\[ C = \epsilon_0 \frac{S}{d} \][/tex]
[tex]\[ C = 8.854187817 \times 10^{-12} \frac{2}{5 \times 10^{-3}} \][/tex]
[tex]\[ C \approx 3.542 \times 10^{-9} \text{ F} \][/tex]
The capacitance is approximately [tex]\( 3.5416751267999994 \times 10^{-9} \)[/tex] F.
3. (b) Charge on each plate:
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = 3.542 \times 10^{-9} \times 10,000 \][/tex]
[tex]\[ Q \approx 3.542 \times 10^{-5} \text{ C} \][/tex]
The charge on each plate is approximately [tex]\( 3.541675126799999 \times 10^{-5} \)[/tex] C.
4. (c) Electric field magnitude:
[tex]\[ E = \frac{V}{d} \][/tex]
[tex]\[ E = \frac{10,000}{5 \times 10^{-3}} \][/tex]
[tex]\[ E = 2,000,000 \text{ V/m} \][/tex]
The magnitude of the electric field is approximately [tex]\( 2,000,000.0 \)[/tex] V/m.
### Part (c)
Given:
- Capacitance [tex]\( C_1 = 6 \)[/tex] µF
- Capacitance [tex]\( C_2 = 3 \)[/tex] µF
- Voltage [tex]\( V \)[/tex] = 18 V
Solution:
i) In Series:
1. Equivalent capacitance:
[tex]\[ \frac{1}{C_\text{series}} = \frac{1}{C_1} + \frac{1}{C_2} \][/tex]
[tex]\[ \frac{1}{C_\text{series}} = \frac{1}{6 \times 10^{-6}} + \frac{1}{3 \times 10^{-6}} \][/tex]
[tex]\[ C_\text{series} = 2 \times 10^{-6} \text{ F} \][/tex]
The equivalent capacitance in series is [tex]\( 2 \times 10^{-6} \)[/tex] F (or 2 µF).
2. Charge in series (same for both):
[tex]\[ Q_\text{series} = C_\text{series} \times V \][/tex]
[tex]\[ Q_\text{series} = 2 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_\text{series} = 3.6 \times 10^{-5} \text{ C} \][/tex]
The charge in series is [tex]\( 3.6 \times 10^{-5} \)[/tex] C.
3. Voltage across each capacitor:
[tex]\[ V_{C1(\text{series})} = \frac{Q_\text{series}}{C_1} \][/tex]
[tex]\[ V_{C1(\text{series})} = \frac{3.6 \times 10^{-5}}{6 \times 10^{-6}} \][/tex]
[tex]\[ V_{C1(\text{series})} = 6 \text{ V} \][/tex]
[tex]\[ V_{C2(\text{series})} = \frac{Q_\text{series}}{C_2} \][/tex]
[tex]\[ V_{C2(\text{series})} = \frac{3.6 \times 10^{-5}}{3 \times 10^{-6}} \][/tex]
[tex]\[ V_{C2(\text{series})} = 12 \text{ V} \][/tex]
The voltage across [tex]\( C_1 \)[/tex] is 6 V, and across [tex]\( C_2 \)[/tex] is 12 V.
ii) In Parallel:
1. Equivalent capacitance:
[tex]\[ C_\text{parallel} = C_1 + C_2 \][/tex]
[tex]\[ C_\text{parallel} = 6 \times 10^{-6} + 3 \times 10^{-6} \][/tex]
[tex]\[ C_\text{parallel} = 9 \times 10^{-6} \text{ F} \][/tex]
The equivalent capacitance in parallel is [tex]\( 9 \times 10^{-6} \)[/tex] F (or 9 µF).
2. Charge on each capacitor:
[tex]\[ Q_{C1(\text{parallel})} = C_1 \times V \][/tex]
[tex]\[ Q_{C1(\text{parallel})} = 6 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_{C1(\text{parallel})} = 108 \times 10^{-6} \text{ C} \][/tex]
The charge on [tex]\( C_1 \)[/tex] is [tex]\( 1.08 \times 10^{-4} \)[/tex] C (or 108 µC).
[tex]\[ Q_{C2(\text{parallel})} = C_2 \times V \][/tex]
[tex]\[ Q_{C2(\text{parallel})} = 3 \times 10^{-6} \times 18 \][/tex]
[tex]\[ Q_{C2(\text{parallel})} = 54 \times 10^{-6} \text{ C} \][/tex]
The charge on [tex]\( C_2 \)[/tex] is [tex]\( 5.4 \times 10^{-5} \)[/tex] C (or 54 µC).
Summary:
- (a) Capacitance: [tex]\(\approx 4.427 \times 10^{-11} \text{ F}\)[/tex]
- (b) Capacitance: [tex]\(\approx 3.542 \times 10^{-9} \text{ F}\)[/tex]
- Charge: [tex]\(\approx 3.542 \times 10^{-5} \text{ C}\)[/tex]
- Electric field: [tex]\(\approx 2,000,000.0 \text{ V/m}\)[/tex]
- (c-i)
- Series Capacitance: [tex]\(2 \times 10^{-6} \text{ F}\)[/tex]
- Charge: [tex]\(3.6 \times 10^{-5} \text{ C}\)[/tex]
- Voltage across [tex]\( C_1 \)[/tex]: 6 V
- Voltage across [tex]\( C_2 \)[/tex]: 12 V
- (c-ii)
- Parallel Capacitance: [tex]\(9 \times 10^{-6} \text{ F}\)[/tex]
- Charge on [tex]\( C_1 \)[/tex]: [tex]\(1.08 \times 10^{-4} \text{ C}\)[/tex]
- Charge on [tex]\( C_2 \)[/tex]: [tex]\(5.4 \times 10^{-5} \text{ C}\)[/tex]
This completes the detailed step-by-step solution for each part of the question.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
