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parallel plate capacitor has capacitance `C` when no dielectric between thw plates. Now a slab of dielectric constant `K`, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. the new capacitance will be

A

`(K + 1) (C )/(4)`

B

`(K + 2) (C )/(4)`

C

`(K +3) (C )/(4)`

D

`(KC)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
C
We can observe that system is equivalent to two capacitors `C_(1)` and `C_(2)` in parallel.

Capacitance of original capacitor without dipole.
`C_(1) = (epsilon_(0))/(d) ((3A)/(4)) = (3 epsilon_(0)A)/(4d)`
With medium, `C_(2) = (epsilon_(0)K)/(d) ((A)/(4)) = (epsilon_(0)AK)/(4d)`
`C' = C_(1) + C_(2)`
Or `C' = (3epsilon_(0)A)/(4d) + (epsilon_(0)AK)/(4d) = (epsilon_(0)A)/(d) [(3)/(4)+ (K)/(4)]`
Or `C' = (C)/(4) (K + 3)` [`:' C = (A epsilon_(0))/(d)]`
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Knowledge Check

  • A parallel plate capacitor with vacuum between its plates has capacitance C . A slabe of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill 1//3^(rd) of the capacitor as shown in the figure. the new capacitance will be

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