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A parallel plate capacitor with square p...

A parallel plate capacitor with square plates is filed with four dielectrics of dielectric constants `K_(1),K_(2),K_(3),K_(4),` arranged as shown in the figure. The effective dielectric constant K will be :

A

`K=((K_(1)+K_(4))(K_(2)+K_(3)))/(2(K_(1)+K_(2)+K_(3)+K_(4)))`

B

`K=[(K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(4))/(K_(2)+K_(4))]`

C

`K=((K_(1)+K_(2))(K_(3)+K_(4)))/(2(K_(1)+K_(2)+K_(3)+K_(4)))`

D

`K=((K_(1)+K_(2))(K_(3)+K_(4)))/((K_(1)+K_(2)+K_(3)+K_(4)))`

Text Solution

Verified by Experts

The correct Answer is:
B

`C_(1)=(K_(1)(A)/(2)epsilon_(0))/((d)/(2))=(K_(1)Aepsilon_(0))/(d)`
`C_(2)=(K_(2)Aepsilon_(0))/(d)`
`C_(3)=(K_(3)Aepsilon_(0))/(d)`
`C_(4)=(K_(4)Aepsilon_(0))/(d)`
`C_(eq)=(C_(1)C_(3))/(C_(1)+C_(3))+(C_(2)C_(4))/(C_(2)+C_(4))`
`C_(eq)=K_(eq)(Aepsilon_(0))/(d)=[(K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(4))/(K_(2)+K_(4))](Aepsilon_(0))/(d)`
`K_(eq)=[(K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(4))/(K_(2)+K_(4))]`
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