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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'.
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A parallel plate capacitor with air between plates has capacity of 1 ...
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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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