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A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to
- a)6 × 104 C/m2
- b)6 × 10-7 C/m2
- c)3 × 10-7 C/m2
- d)3 × 104 C/m2
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A parallel plate capacitor is made of two circular plates separated b...
Here, dielectric constant, K = 2.2 and E = 3 × 104 V m-1
Electric field between the parallel plate capacitor with dielectric, E = σ / Kε0
So, charge density, σ = Kε0E = 2.2 × 8.85 × 10-12 × 3 × 104 = 5.8 × 10-7 or 6 × 10-7 C m-2.
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A parallel plate capacitor is made of two circular plates separated b...
Concept:
The electric field between two plates of a parallel plate capacitor is given by:
E = σ/ε₀
where σ is the surface charge density on the plates, and ε₀ is the permittivity of free space. If a dielectric material is placed between the plates, the electric field is reduced by a factor of the dielectric constant of the material.
E = σ/(ε₀εᵣ)
where εᵣ is the relative permittivity or dielectric constant of the material.
Calculation:
Given: distance between the plates (d) = 5 mm = 0.005 m, electric field in the dielectric (E) = 3 × 10^4 V/m, dielectric constant (εᵣ) = 2.2
We know that the electric field between the plates of a parallel plate capacitor is given by:
E = σ/(ε₀εᵣ)
Rearranging the equation, we get:
σ = Eε₀εᵣ
Substituting the given values, we get:
σ = (3 × 10^4)(8.85 × 10^-12)(2.2)
σ = 5.934 × 10^-7 C/m^2
Therefore, the charge density of the positive plate is approximately 6 × 10^-7 C/m^2, which is option B.
The electric field between two plates of a parallel plate capacitor is given by:
E = σ/ε₀
where σ is the surface charge density on the plates, and ε₀ is the permittivity of free space. If a dielectric material is placed between the plates, the electric field is reduced by a factor of the dielectric constant of the material.
E = σ/(ε₀εᵣ)
where εᵣ is the relative permittivity or dielectric constant of the material.
Calculation:
Given: distance between the plates (d) = 5 mm = 0.005 m, electric field in the dielectric (E) = 3 × 10^4 V/m, dielectric constant (εᵣ) = 2.2
We know that the electric field between the plates of a parallel plate capacitor is given by:
E = σ/(ε₀εᵣ)
Rearranging the equation, we get:
σ = Eε₀εᵣ
Substituting the given values, we get:
σ = (3 × 10^4)(8.85 × 10^-12)(2.2)
σ = 5.934 × 10^-7 C/m^2
Therefore, the charge density of the positive plate is approximately 6 × 10^-7 C/m^2, which is option B.
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A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close toa)6 × 104 C/m2b)6 × 10-7 C/m2c)3 × 10-7 C/m2d)3 × 104 C/m2Correct answer is option 'B'. Can you explain this answer?
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A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close toa)6 × 104 C/m2b)6 × 10-7 C/m2c)3 × 10-7 C/m2d)3 × 104 C/m2Correct answer is option 'B'. 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 is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close toa)6 × 104 C/m2b)6 × 10-7 C/m2c)3 × 10-7 C/m2d)3 × 104 C/m2Correct answer is option 'B'. 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 is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close toa)6 × 104 C/m2b)6 × 10-7 C/m2c)3 × 10-7 C/m2d)3 × 104 C/m2Correct answer is option 'B'. Can you explain this answer?.
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