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A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N is
Correct answer is '8'. Can you explain this answer?
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A parallel plate capacitor of capacitance 100 pF is to be constructed...
Explanation:
Given, Capacitance of the parallel plate capacitor, C = 100 pF and thickness of paper sheet, t = 1 mm = 0.1 cm.
Dielectric constant of paper, εr = 4.0 and diameter of the circular metal foils, d = 2.0 cm.
Using the formula for capacitance of a parallel plate capacitor with dielectric medium, we have:
C = ε0 * εr * A / t
where ε0 is the permittivity of free space and A is the area of the plates.
Therefore, A = C * t / ε0 * εr
Substituting the given values, we get:
A = 100 * 10^-12 * 0.1 / (8.85 * 10^-12 * 4.0) = 0.283 m^2
Area of each metal foil, a = πd^2 / 4 = π * (2.0)^2 / 4 = 3.14 cm^2
Number of metal foils required, N = A / a = 0.283 / 0.0314 ≈ 9
Therefore, the number of circular metal foils of diameter 2.0 cm is 8 (N = 8) because we cannot use a fractional number of metal foils.
Given, Capacitance of the parallel plate capacitor, C = 100 pF and thickness of paper sheet, t = 1 mm = 0.1 cm.
Dielectric constant of paper, εr = 4.0 and diameter of the circular metal foils, d = 2.0 cm.
Using the formula for capacitance of a parallel plate capacitor with dielectric medium, we have:
C = ε0 * εr * A / t
where ε0 is the permittivity of free space and A is the area of the plates.
Therefore, A = C * t / ε0 * εr
Substituting the given values, we get:
A = 100 * 10^-12 * 0.1 / (8.85 * 10^-12 * 4.0) = 0.283 m^2
Area of each metal foil, a = πd^2 / 4 = π * (2.0)^2 / 4 = 3.14 cm^2
Number of metal foils required, N = A / a = 0.283 / 0.0314 ≈ 9
Therefore, the number of circular metal foils of diameter 2.0 cm is 8 (N = 8) because we cannot use a fractional number of metal foils.
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Community Answer
A parallel plate capacitor of capacitance 100 pF is to be constructed...
Let the number of sheets required be 'n'.
They will form (n - 1) capacitors in series. If K is the dielectric constant of dielectric, the capacitance is given by:
So,
(n - 1) = 9
n = 10
The number of circular metal foils of diameter 2.0 cm is 10.
N + 2 = 10 or N = 8
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A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer?
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A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer?.
A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer?.
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A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer?, a detailed solution for A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? has been provided alongside types of A parallel plate capacitor of capacitance 100 pF is to be constructed using paper sheet of 1 mm thickness as dielectric. If the dielectric constant of paper is 4.0, then the number of circular metal foils of diameter 2.0 cm is (N + 2). The value of N isCorrect answer is '8'. Can you explain this answer? theory, EduRev gives you an
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