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Question
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
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Solution
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, the distance between the parallel plates was d and it was filled with air. The dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
`"C" = ("k"in_0"A")/"d"`
= `(in_0"A")/"d"` .......(i)
Where,
A = Area of each plate
`in_0` = Permittivity of free space
If the distance between the plates is reduced to half, then the new distance, d’ = `"d"/2`
Dielectric constant of the substance filled in between the plates, k' = 6
Hence, the capacitance of the capacitor becomes
`"C'" = ("k'"in_0"A")/"d"`
= `(6in_0"A")/("d"/2)` ........(ii)
Taking ratios of equations (i) and (ii), we obtain
C' = 2 × 6 C
= 12 C
= 12 × 8
= 96 pF
Therefore, the capacitance between the plates is 96 pF.
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