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A dielectric of thickness 5 cm and dielectric constant 10 is introduced in between the plates of a parallel plate capacitor having plate area 500 sq.cm and separation between plates 10 cm. The capacitance of the capacitor is ∈o = 8.8x10-12  SI unit
  • a)
    8pF
  • b)
    6pF
  • c)
    4pF
  • d)
    20pF
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A dielectric of thickness 5 cm and dielectric constant 10 is introduce...
Given data:
- Thickness of dielectric (d) = 5 cm
- Dielectric constant (k) = 10
- Plate area (A) = 500 sq.cm
- Separation between plates (d) = 10 cm
- Permittivity of free space (ε0) = 8.8x10-12 F/m

Formula for capacitance of a parallel plate capacitor:
C = (ε0 * k * A) / d

Calculating capacitance without dielectric:
C0 = (ε0 * A) / d
C0 = (8.8x10-12 * 500) / 10
C0 = 4.4x10-10 F

Calculating capacitance with dielectric:
C = (k * C0)
C = 10 * 4.4x10-10
C = 4.4x10-9 F
C = 4.4 pF
Therefore, the capacitance of the capacitor with the dielectric is 4.4 pF, which is the correct answer.
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A dielectric of thickness 5 cm and dielectric constant 10 is introduced in between the plates of a parallel plate capacitor having plate area 500 sq.cm and separation between plates 10 cm. The capacitance of the capacitor is ∈o= 8.8x10-12 SI unita)8pFb)6pFc)4pFd)20pFCorrect answer is option 'A'. Can you explain this answer?
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