The plates of a paraller-plate capacitor...

The plates of a paraller-plate capacitor are made of circular discs of radii` 5.0`cm each .If the separation between the plates is `1.0`mm,What is the capacitance?

Text Solution

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Area of the plate
`A = pie r^(2) = pie xx (5 xx 10^(-2))^(2)`
`d = 1.0 xx 10^(-3) m `
`C = eplision_0 A / d`
`= 8.8 xx 10^(-12) xx 3.14 xx 25 xx 10^(-4) / 10^(-3)`
`= 695 xx 10 ^(-5) mu F`

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A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq. cm and separation between the plates 10cm. The capacitance of the capacitor with dielectric slab is (epsi_(0) = 8.8 xx 10^(-12)C^(2)//N-m^(2))

4.4pF

6.2pF

8pF

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The plate area of a parallel-plate capacitors is 10cm^(2) and its capacitance is 2pF . The separation between the plates of the capacitors is close to (in mm). (permittivity of vacuum, epsi_(0)=(80)/(9)xx10^(-12)m^(-3)kg^(-1)s^(4)A^(2),1pF=10^(-12)F )

2.2

4.4

8.8

17.6

The total capacitance if the parallel plate capacitor has n number of interleaved plates, area A and separation d between them,

`(epsi _(0) n A)/( d )`

`(epsi _(0) A )/( d)`

`(epsi _(0) (n -1) A)/( d)`

`(epsi _(0) n ^(2)A)/( d)`