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a) A parallel-plate capacitor has an effective area [tex]\( S = 25 \, \text{cm}^2 \)[/tex], a plate separation [tex]\( d = 0.5 \, \text{mm} \)[/tex], and an air dielectric. What is its capacitance?

b) The plates of a parallel-plate capacitor in a vacuum are [tex]\( 5 \, \text{mm} \)[/tex] apart and have an area of [tex]\( 2 \, \text{m}^2 \)[/tex]. A potential difference of [tex]\( 10,000 \, \text{V} \)[/tex] (10 kV) is applied across the capacitor. Compute:
1. The capacitance
2. The charge on each plate
3. The magnitude of the electric field in the space between them.

c) Consider two capacitors, [tex]\( C_1 = 6 \, \mu\text{F} \)[/tex] and [tex]\( C_2 = 3 \, \mu\text{F} \)[/tex], connected to an 18 V DC power supply. Find the equivalent capacitance, charge, and potential difference for each capacitor when the two capacitors are connected:
1. In series.
2. In parallel.



Answer :

GinnyAnswer
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.

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