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A parallel plate capacitor with air between plates has capacity of 1 pF. Now plates are moved apart with insulating handle so that separation is twice of original value. The space between plates is filled with a medium of dielectric constant 6. The new value of capacity in pF is
- a)2
- b)3
- c)1
- d)6
Correct answer is option 'B'. Can you explain this answer?
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A parallel plate capacitor with air between plates has capacity of 1 ...
Given information:
- A parallel plate capacitor with air between the plates has a capacity of 1 pF.
- The plates are moved apart with an insulating handle, doubling the separation between them.
- The space between the plates is then filled with a medium of dielectric constant 6.
To find: The new value of capacity in pF.
Solution:
1. Capacitance in 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, A is the area of the plates, and d is the separation between the plates.
2. Initially, the capacitance (C₁) with air between the plates is 1 pF.
3. When the plates are moved apart, the separation between them becomes twice the initial value (2d).
4. Now, the space between the plates is filled with a dielectric medium with a dielectric constant (κ) of 6.
5. The new capacitance (C₂) can be calculated using the formula:
C₂ = κ * C₁
where C₁ is the initial capacitance with air and κ is the dielectric constant of the medium.
6. Substituting the given values, we have:
C₂ = 6 * 1 pF
C₂ = 6 pF
Answer: The new value of capacity is 6 pF, which corresponds to option 'B'.
- A parallel plate capacitor with air between the plates has a capacity of 1 pF.
- The plates are moved apart with an insulating handle, doubling the separation between them.
- The space between the plates is then filled with a medium of dielectric constant 6.
To find: The new value of capacity in pF.
Solution:
1. Capacitance in 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, A is the area of the plates, and d is the separation between the plates.
2. Initially, the capacitance (C₁) with air between the plates is 1 pF.
3. When the plates are moved apart, the separation between them becomes twice the initial value (2d).
4. Now, the space between the plates is filled with a dielectric medium with a dielectric constant (κ) of 6.
5. The new capacitance (C₂) can be calculated using the formula:
C₂ = κ * C₁
where C₁ is the initial capacitance with air and κ is the dielectric constant of the medium.
6. Substituting the given values, we have:
C₂ = 6 * 1 pF
C₂ = 6 pF
Answer: The new value of capacity is 6 pF, which corresponds to option 'B'.
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A parallel plate capacitor with air between plates has capacity of 1 pF. Now plates are moved apart with insulating handle so that separation is twice of original value. The space between plates is filled with a medium of dielectric constant 6. The new value of capacity in pF isa)2b)3c)1d)6Correct answer is option 'B'. Can you explain this answer?
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A parallel plate capacitor with air between plates has capacity of 1 pF. Now plates are moved apart with insulating handle so that separation is twice of original value. The space between plates is filled with a medium of dielectric constant 6. The new value of capacity in pF isa)2b)3c)1d)6Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A parallel plate capacitor with air between plates has capacity of 1 pF. Now plates are moved apart with insulating handle so that separation is twice of original value. The space between plates is filled with a medium of dielectric constant 6. The new value of capacity in pF isa)2b)3c)1d)6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor with air between plates has capacity of 1 pF. Now plates are moved apart with insulating handle so that separation is twice of original value. The space between plates is filled with a medium of dielectric constant 6. The new value of capacity in pF isa)2b)3c)1d)6Correct answer is option 'B'. Can you explain this answer?.
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