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A parallel -plate capacitor of area A, p...

A parallel -plate capacitor of area `A`, plate separation `d` and capacitance `C` is filled with four dielectric materials having dielectric constant `k_(1), k_(2), k_(3)` and `k_(4)` as shown in the figure below. If a single dielectric materical is to be used to have the same capacitance `C` in this capacitor, then its dielectric constant `k` is given by

A

`(2)/(k)= (3)/(k_(1) + k_(2) + k_(3)) + (1)/(k_(4))`

B

`(1)/(k)= (1)/(k_(1)) + (1)/(k_(2)) + (1)/(k_(1))= (3)/(2k_(4))`

C

`k= k_(1) + k_(2) + k_() + 3k_(4)`

D

`k=(2)/(3) (k_(1) + k_(2) + k_(3)) + 2k_(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`C_(1)= (2k_(1) epsi_(0)A)/(3d) C_(2)= (2k_(2) epsi_(0)A)/(3d)`
`C_(3)= (2k_(1) epsi_(0)A)/(3d) C_(4)= (k_(1)2 epsi_(0)A)/(d)`
`(1)/(C )= (1)/(C_(1) + C_(2) +C_(3)) + (1)/(C_(4))`
solving we get
`C= (k epsi_(0)A)/(d)`
`therefore (d)/(k epsi_(0)A) = (d)/(2 epsi_(0)A) [(3)/(k_(1) +K_(2) + K_(3))+ (1)/(k_(4))]`
`rArr (2)/(K)= [(3)/(k_(1)+ k_(2) + k_(3)) + (1)/(k_(4))]`
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