A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes pF. The value of is __________.
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Answer: 105
Solution
Original capacitance C = ε₀A/d = 15 pF.
New capacitance: separation doubles (d → 2d) and dielectric constant K = 3.5 is inserted.
C' = Kε₀A/(2d) = (3.5/2) × (ε₀A/d) = (3.5/2) × 15 = 52.5/2 = 26.25 pF.
Setting this equal to x/4: x/4 = 26.25, so x = 105.
New capacitance: separation doubles (d → 2d) and dielectric constant K = 3.5 is inserted.
C' = Kε₀A/(2d) = (3.5/2) × (ε₀A/d) = (3.5/2) × 15 = 52.5/2 = 26.25 pF.
Setting this equal to x/4: x/4 = 26.25, so x = 105.
[JEE Main 2023 · 24 Jan · Shift 2]