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Parallel plate capacitor has capacitance...

Parallel plate capacitor has capacitance `C_(0) = (epsilon_(0)A)/(d)`. When a insulating medium is added of dielectric constant K, the capacity becomes `C_(m) = (epsilon A)/(d) = K(epsilon_(0)A)/(d) = K C_(0)`. Let t be the thickness of dielectric such that :
`t( lt d)`, then the capacity is `C_(d) = (epsilon_(0)A)/(d-t(1-(1)/(k)))`. For metals, `K = oo`. Therefore, when a metal plate of thickness `t lt d` is inctroduced, the capacity becomes `C = (epsilon_(0)A)/(d-t)`.

In a capacitor, a metal plate of thickness `t = (d)/(2)` is introduced. The increase in its capacity is :

A

`50%`

B

`100%`

C

`200%`

D

`1%`

Text Solution

Verified by Experts

The correct Answer is:
B
`C._(0) = (epsilon_(0)A)/(d-t) = (epsilon_(0)A)/(d-(d)/(2)) = (2 epsilon_(0)A)/(d) = 2C_(0)`
Increase in capacity
`= C. = C_(0) = 2 C_(0) - C_(0) = C_(0)`
Percentage increase in capacity
`= (C_(0))/(C_(0)) xx 100 = 100%`
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