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A dielectric slab of thickness 1.0 cm a...

A dielectric slab of thickness 1.0 cm and dielectric constant 5 is placed between the plates of a parallel plate capacitor of plate area `0.01 m^(2)` and separation 2.0 cm. Calculate the change in capacity on introduction of dielectric. What would be on the change, if the dielectric slab were conducting?

Text Solution

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Here, `t = 1.0 cm = 10^(-2) m`,
`in_(r) = K = 5, A = 0.01 m^(2) = 10^(-2) m^(2)`,
`d = 2 cm = 2xx10^(-2) m`
Capacity with air inbetween the plates
`C_(0) = (in_(0) A)/(d) = (8.85xx10^(-12)xx10^(-2))/(2xx10^(-2))`
`= 4.425xx10^(-12)` farad
Capacity the dielectric slab inbetween the plates
`C = (in_(0) A)/(d-t(1- (1)/(k))) = (8.85xx10^(-12)xx10^(-2))/(2xx10^(-2) - 10^(-2) (1 - (1)/(5)))`
`= 7.375xx10^(-12) farad`
Capacity with coducting slab inbetween the plates
`C' = (in_(0) A)/(d -t) = (8.85xx10^(-12)xx10^(-2))/(2xx10^(-2) -1xx10^(-2))`
`C' = (8.85xx10^(-14))/(10^(-2)) = 8.85xx10^(-12)` farad.
Increate in capacity on introduction of dielectric.
`= C - C_(0) = 7.735xx10^(-12) - 4.425xx10^(-12)`
`= 2.95xx10^(-12)` farad
Increase in capacity on indrocuction of conducting slab
`= C' - C_(0) = 8.85xx10^(-12) - 4.425xx10^(-12)`
`= 4.425xx10^(-12)` farad
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