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A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?
Correct answer is '11.4'. Can you explain this answer?
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A parallel plate capacitor with plate area 40 cm2 and plate separation...
The electric field due to the polarization charges is
The electric field inside the dielectric slab is
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A parallel plate capacitor with plate area 40 cm2 and plate separation...
Given:
- Plate area (A) = 40 cm²
- Plate separation (d) = 5 mm = 0.5 cm
- Charge (Q) = 2 * 10^(-12) C
- Dielectric constant (k) = 5
- Dielectric thickness (t) = 3 mm = 0.3 cm
To find:
- Capacitance (C)
Formula:
The capacitance of a parallel plate capacitor with a dielectric is given by the formula:
C = (k * ε₀ * A) / d
where ε₀ is the permittivity of free space (ε₀ ≈ 8.854 * 10^(-12) F/m)
Solution:
Given that the plate area is 40 cm², we can convert it to m²:
A = 40 cm² = 40 * 10^(-4) m²
Similarly, the plate separation is 0.5 cm, which can be converted to meters:
d = 0.5 cm = 0.5 * 10^(-2) m
Now, let's calculate the capacitance without the dielectric:
C₁ = (ε₀ * A) / d
The permittivity of free space is approximately ε₀ ≈ 8.854 * 10^(-12) F/m, so substituting the values:
C₁ = (8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₁ = (8.854 * 40) / 0.5
C₁ = 708.32 F
Next, we need to calculate the capacitance with the dielectric:
C₂ = (k * ε₀ * A) / d
Substituting the values:
C₂ = (5 * 8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₂ = (5 * 8.854 * 40) / 0.5
C₂ = 1770.8 F
The capacitance of the capacitor with the dielectric is given by:
C = C₂ - C₁
C = 1770.8 F - 708.32 F
C ≈ 1062.48 F
However, the question asks for the capacitance in units of '11.4'. Therefore, the given answer seems to be incorrect. Please double-check the answer.
- Plate area (A) = 40 cm²
- Plate separation (d) = 5 mm = 0.5 cm
- Charge (Q) = 2 * 10^(-12) C
- Dielectric constant (k) = 5
- Dielectric thickness (t) = 3 mm = 0.3 cm
To find:
- Capacitance (C)
Formula:
The capacitance of a parallel plate capacitor with a dielectric is given by the formula:
C = (k * ε₀ * A) / d
where ε₀ is the permittivity of free space (ε₀ ≈ 8.854 * 10^(-12) F/m)
Solution:
Given that the plate area is 40 cm², we can convert it to m²:
A = 40 cm² = 40 * 10^(-4) m²
Similarly, the plate separation is 0.5 cm, which can be converted to meters:
d = 0.5 cm = 0.5 * 10^(-2) m
Now, let's calculate the capacitance without the dielectric:
C₁ = (ε₀ * A) / d
The permittivity of free space is approximately ε₀ ≈ 8.854 * 10^(-12) F/m, so substituting the values:
C₁ = (8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₁ = (8.854 * 40) / 0.5
C₁ = 708.32 F
Next, we need to calculate the capacitance with the dielectric:
C₂ = (k * ε₀ * A) / d
Substituting the values:
C₂ = (5 * 8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₂ = (5 * 8.854 * 40) / 0.5
C₂ = 1770.8 F
The capacitance of the capacitor with the dielectric is given by:
C = C₂ - C₁
C = 1770.8 F - 708.32 F
C ≈ 1062.48 F
However, the question asks for the capacitance in units of '11.4'. Therefore, the given answer seems to be incorrect. Please double-check the answer.
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A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer?
Question Description
A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer?.
A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor with plate area 40 cm2 and plate separation 5 mm is given a charge Q = 2 * 10-12C. The battery is disconnected and a parallel faced dielectric slab (k = 5) of area 40 cm2 and thickness 3 mm is placed centrally between the two plates. What is the capacitance?Correct answer is '11.4'. Can you explain this answer?.
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