Three parallel plate air capacitors are ...

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area `A/3`and the separation between the plates is d, 2d and 3d respectively. The equivalent capacity of combination is `(epsilon_0` = absolute permittivity of free space)

`(7in_(0)A)/(18d)`

`(11in_(0)A)/(18d)`

`(13in_(0)A)/(18d)`

`(17in_(0)A)/(18d)`

Text Solution

Verified by Experts

The correct Answer is:

`C_(p)=C_(1)+C_(2)+C_(3)`
`=(A_(1)epsi_(0))/(d_(1))+(A_(2)epsi_(0))/(d_(2))+(A_(3)epsi_(0))/(d_(3))`
`=(Aepsi_(0))/(3d)+(Aepsi_(0))/(6d)+(Aepsi_(0))/(9d)`
`=(Aepsi_(0))/(3d)(1+(1)/(2)+(1)/(3))=(Aepsi_(0))/(3d)=((6+3+2)/(6))`
`=(11)/(18)(Aepsi_(0))/(d)`.

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Three parallel plate air capacitors are connected in parallel each capacitor has plate area ('A')/(3) and the separation between the plates is 'd', '2d' and '3d' respectively. The equivalaent capacity of combination is ( epsi_(0) =absolute permittivity of free space).

`(7epsi_(0)A)/(18d)`

`(11epsi_(0)A)/(18d)`

`(13epsi_(0)A)/(18d)`

`(17epsi_(0)A)/(18d)`

Area of a parallel plate capacitor of capacitance 2F and separation between the plates 0.5cm will be

`1.13xx10^(9) m^(2)`

`1.13xx10^(6) m^(2)`

`10^(8) m^(2)`

`1.13 m^(2)`

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)`