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)

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

`(1epsilon_(0)A)/(18d`

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

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

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The correct Answer is:

According to question, capacitance of parallel plate
`"capacitor is given by C"=(epsilon_(0)A)/(d)`
`"For first capacitor, "C_(1)=(epsilon_(0)A)/(3d)`
`"For second capacitor, "C_(2)=(epsilon_(0)A)/(6d)`
`"For third capacitor," C_(3)=(epsilon_(0)A)/(9d)`
Again they are arranged in parallel combination, so equivalent capacitor is given by
`C_("eq")=C_(1)+C_(2)+C_(3)`
`=(epsilon_(0)A)/(d)xx(11)/(18)=(11epsilon_(0)A)/(18d)`

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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)`