A parallel plate air capacitor has capacity C and plate area A . Now the space between its plates is filled by two slabs of thickness 2 d each, having dielectric constants K 1 and K 2 . Distance

Question

A parallel plate air capacitor has capacity C and plate area A . Now the space between its plates is filled by two slabs of thickness 2 d ​ each, having dielectric constants K 1 ​ and K 2 ​ . Distance of separation between plates is d . The capacity of the capacitor is ( ε 0 ​ = permittivity of free space)

TestHub · Accepted Answer

Initial capacitance of the air capacitor is $C = \frac{ε _{0} A}{d}$

When two dielectric slabs of thickness $\frac{d}{2}$ are introduced between the plates, they act as two capacitors connected in series.

Capacitance of the first part is $C_{1} = \frac{K _{1} ε _{0} A}{d /2} = \frac{2 K _{1} ε _{0} A}{d} = 2 K_{1} C$

Capacitance of the second part is $C_{2} = \frac{K _{2} ε _{0} A}{d /2} = \frac{2 K _{2} ε _{0} A}{d} = 2 K_{2} C$

The equivalent capacitance $C_{e q}$ for the series combination is given by:

$C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$

$C_{e q} = \frac{( 2 K _{1} C ) ( 2 K _{2} C )}{2 K _{1} C + 2 K _{2} C}$

$C_{e q} = \frac{4 K _{1} K _{2} C ^{2}}{2 C ( K _{1} + K _{2} )}$

$C_{e q} = 2 C (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

Answer: $2 C (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

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