A parallel plate capacitor has plate area A and plate separation d. The space between the plates is filled up to a thickness `x lt d` with a dielectric constant K. Calculate the capacitance of the system.
A
`(2Kepsilon_A)/(Kd - x(K-1)`
B
`(Kepsilon_A)/(K2d - x(K-1)`
C
`(Kepsilon_A)/(K - x(K-1)`
D
`(Kepsilon_A)/(Kd - x(K-1)`
Text Solution
Verified by Experts
The correct Answer is:
D
The situation is shown in . The given system is equivalent to the series combinaton of two capacitors, one between a and c and the other between c and b. Here c represents the upper surface of the dielectric. This is because the potential at the upper surface of the dielectric is constant and we can imagine a thin metal plate being placed there. The capacitance of the capacitor between a and c is ` `C_1 = (Kepsilon_o A)/(x)` ` and the between c and b is ` `C_2 = (epsilon_0A)/(d-x)` `. The equivalent capacitance is ` `C=(C_1C_2)/(C_1 +C_2) = (Kepsilion_A)/(Kd - x(K-1)` `.
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