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?
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
Where, A = Area of each plate
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, d' = d / 2
Dielectric constant of the substance filled in between the plates, = 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF.
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(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
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Would you expect a greater or smaller dip angle in Britain?
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(b) (E g ) C < (E g ) Ge > (E g ) Si
(c) (E g ) C > (E g ) Si > (E g ) Ge
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The capacitance of a parallel plate capacitor having air between its plates is 10 µf if the distance between the plates is reduced to half and the medium between plates is replaced by a medium of dielectric constant k=5, then find the value how much increase or decrease in the capacitance of the capacitor.
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Thnk u very helpful n good answer
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Akansha