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A parallel plate capacitor of area A, pl...

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

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
Applying `C=(epsilon_0A)/(d-t_1-t_2+(t_1)/(K_1)+(t_2)/(K_2))` we have
`(epsilon_0(A//2))/(d-d//2-d//2+(d//2)/(K_1)+(d//2)/(K_3))+(epsilon_0(A//2))/(d-d//2-d//2+(d//2)/(K_2)+(d//2)/(K_3))=(Kepsilon_0A)/d`
Solving the equation we get ,`K=(K_1K_3)/(K_1+K_3)+(K_2K_3)/(K_2+K_3)`
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