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The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ?
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Introduction:
In this problem, we are given a parallel-plate capacitor with two metallic plates of radius r and distance d apart, having a capacity C. We have to find the new capacity of the capacitor when a plate of radius r/2 and thickness d with a dielectric constant of 6 is placed between the plates.
Solution:
We can use the formula for capacitance of a parallel-plate capacitor with a dielectric material between the plates:
C' = (εA) / (d + t)
Where,
C' = new capacitance of the capacitor,
ε = dielectric constant of the material,
A = area of the plates,
d = distance between the plates, and
t = thickness of the dielectric material.
Calculation:
Let us calculate the new capacitance:
Given, radius of plates, r
Radius of dielectric plate, r/2
Distance between plates, d
Thickness of dielectric plate, d
Dielectric constant, ε = 6
Area of each plate, A = πr²
Area of dielectric plate, A' = π(r/2)²
New distance between plates, d' = d - t
New capacitance, C' = (εA'A) / (d - t)
= (6π(r/2)²πr²) / (d - d)
= (3πr²) / 2
Therefore, the new capacitance of the capacitor is (3πr²) / 2.
Conclusion:
The addition of a dielectric material between the plates of a parallel-plate capacitor increases its capacitance. In this problem, the capacitance of the capacitor increases from C to (3πr²) / 2 when a plate of radius r/2 and thickness d with a dielectric constant of 6 is placed between the plates.
In this problem, we are given a parallel-plate capacitor with two metallic plates of radius r and distance d apart, having a capacity C. We have to find the new capacity of the capacitor when a plate of radius r/2 and thickness d with a dielectric constant of 6 is placed between the plates.
Solution:
We can use the formula for capacitance of a parallel-plate capacitor with a dielectric material between the plates:
C' = (εA) / (d + t)
Where,
C' = new capacitance of the capacitor,
ε = dielectric constant of the material,
A = area of the plates,
d = distance between the plates, and
t = thickness of the dielectric material.
Calculation:
Let us calculate the new capacitance:
Given, radius of plates, r
Radius of dielectric plate, r/2
Distance between plates, d
Thickness of dielectric plate, d
Dielectric constant, ε = 6
Area of each plate, A = πr²
Area of dielectric plate, A' = π(r/2)²
New distance between plates, d' = d - t
New capacitance, C' = (εA'A) / (d - t)
= (6π(r/2)²πr²) / (d - d)
= (3πr²) / 2
Therefore, the new capacitance of the capacitor is (3πr²) / 2.
Conclusion:
The addition of a dielectric material between the plates of a parallel-plate capacitor increases its capacitance. In this problem, the capacitance of the capacitor increases from C to (3πr²) / 2 when a plate of radius r/2 and thickness d with a dielectric constant of 6 is placed between the plates.
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The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ?
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The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ?.
The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The two metallic plates of radius r placed at a distance d apart and it's capacity is C . If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then it's capacity will be ?.
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