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The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).
Correct answer is '2.50 to 2.55'. Can you explain this answer?
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The capacitance of an air-filled parallel-plate capacitor is 60 pF. Wh...
**Given Data:**
- The capacitance of the air-filled parallel-plate capacitor is 60 pF.
- When a dielectric slab is placed on one of the plates, covering it entirely, the capacitance becomes 86 pF.
- The thickness of the dielectric slab is half the distance between the plates.
**To find:**
The relative permittivity of the dielectric.
**Solution:**
The capacitance of a parallel-plate capacitor is given by the formula:
C = (ε₀ * A) / d,
where C is the capacitance, ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m), A is the area of the plates, and d is the distance between the plates.
Let's assume the original distance between the plates is d₀ and the area of the plates is A₀.
From the given information:
C₀ = 60 pF = (ε₀ * A₀) / d₀ …(1)
When the dielectric slab is placed on one of the plates, the new distance between the plates becomes d₁ = d₀/2.
The new capacitance is given as:
C₁ = 86 pF = (ε₁ * A₁) / d₁ …(2)
where ε₁ is the relative permittivity of the dielectric, A₁ is the area of the plates covered by the dielectric, and d₁ is the new distance between the plates.
Now, let's calculate the ratio of the two capacitances:
C₁ / C₀ = (ε₁ * A₁) / d₁ * d₀ / (ε₀ * A₀) …(3)
Since the dielectric slab completely covers one of the plates, the area of the plates covered by the dielectric, A₁, is equal to A₀.
Substituting this in equation (3):
C₁ / C₀ = (ε₁ * A₀) / (d₁ * ε₀ * A₀) * (d₀ / A₀)
= (ε₁ * d₀) / (d₁ * ε₀) …(4)
Substituting the values of C₀ and C₁ from equations (1) and (2) respectively:
60 / 86 = (ε₁ * d₀) / (d₁ * ε₀)
Simplifying:
ε₁ = (60 * d₁ * ε₀) / (86 * d₀)
Given that d₁ = d₀/2, substituting this value:
ε₁ = (60 * (d₀/2) * ε₀) / (86 * d₀)
ε₁ = (30 * ε₀) / 86
ε₁ = (30 * 8.854 × 10⁻¹²) / 86
ε₁ ≈ 3.883 × 10⁻¹² F/m
Converting this value to relative permittivity:
εr = ε₁ / ε₀
εr = (3.883 × 10⁻¹²) / (8.854 × 10⁻¹²)
εr ≈ 0.438
Therefore, the relative permittivity of the dielectric is approximately 0.438.
However, the correct answer given is '2.50 to 2.55'.
This discrepancy may be due to a mistake
- The capacitance of the air-filled parallel-plate capacitor is 60 pF.
- When a dielectric slab is placed on one of the plates, covering it entirely, the capacitance becomes 86 pF.
- The thickness of the dielectric slab is half the distance between the plates.
**To find:**
The relative permittivity of the dielectric.
**Solution:**
The capacitance of a parallel-plate capacitor is given by the formula:
C = (ε₀ * A) / d,
where C is the capacitance, ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m), A is the area of the plates, and d is the distance between the plates.
Let's assume the original distance between the plates is d₀ and the area of the plates is A₀.
From the given information:
C₀ = 60 pF = (ε₀ * A₀) / d₀ …(1)
When the dielectric slab is placed on one of the plates, the new distance between the plates becomes d₁ = d₀/2.
The new capacitance is given as:
C₁ = 86 pF = (ε₁ * A₁) / d₁ …(2)
where ε₁ is the relative permittivity of the dielectric, A₁ is the area of the plates covered by the dielectric, and d₁ is the new distance between the plates.
Now, let's calculate the ratio of the two capacitances:
C₁ / C₀ = (ε₁ * A₁) / d₁ * d₀ / (ε₀ * A₀) …(3)
Since the dielectric slab completely covers one of the plates, the area of the plates covered by the dielectric, A₁, is equal to A₀.
Substituting this in equation (3):
C₁ / C₀ = (ε₁ * A₀) / (d₁ * ε₀ * A₀) * (d₀ / A₀)
= (ε₁ * d₀) / (d₁ * ε₀) …(4)
Substituting the values of C₀ and C₁ from equations (1) and (2) respectively:
60 / 86 = (ε₁ * d₀) / (d₁ * ε₀)
Simplifying:
ε₁ = (60 * d₁ * ε₀) / (86 * d₀)
Given that d₁ = d₀/2, substituting this value:
ε₁ = (60 * (d₀/2) * ε₀) / (86 * d₀)
ε₁ = (30 * ε₀) / 86
ε₁ = (30 * 8.854 × 10⁻¹²) / 86
ε₁ ≈ 3.883 × 10⁻¹² F/m
Converting this value to relative permittivity:
εr = ε₁ / ε₀
εr = (3.883 × 10⁻¹²) / (8.854 × 10⁻¹²)
εr ≈ 0.438
Therefore, the relative permittivity of the dielectric is approximately 0.438.
However, the correct answer given is '2.50 to 2.55'.
This discrepancy may be due to a mistake
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The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).Correct answer is '2.50 to 2.55'. Can you explain this answer?
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The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).Correct answer is '2.50 to 2.55'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).Correct answer is '2.50 to 2.55'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is _____________ (up to 2 decimal places).Correct answer is '2.50 to 2.55'. Can you explain this answer?.
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