A parallel plate capacitor has the space between its plates filled by two slabs of thickness $\frac{d}{2}$ each and dielectric constant $k_{2}$ and d is the plate separation of the capacitor. The capacitance of the capacitor is:
$\frac{2 ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 ε_{0} d}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 ε_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
$\frac{2 ε_{0} A}{d} (K_{1} + K_{2})$

Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness $\frac{d}{2}$ each and dielectric constant $k_{2}$ and d is the plate separation of the capacitor. The capacitance of the capacitor is:
$\frac{2 ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 ε_{0} d}{A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 ε_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
$\frac{2 ε_{0} A}{d} (K_{1} + K_{2})$

Toppr Admin · Accepted Answer

The above effective capacitance is series combination of K1-x3F5-0A2d and K2-x3F5-0A2d-x2234-1Cbf-dK1-x3F5-0A2-dK2-x3F5-0A2-x21D2-Cbf-2-x3F5-0Ad-K1K2K1-K2-

A parallel plate capacitor has the space between its plates filled by two slabs of thickness d2 each and dielectric constant k2 and d is the plate separation of the capacitor. The capacitance of the capacitor is:2ε0Ad(K1+K2K1K2)2ε0dA(K1+K2K1K2)2ε0Ad(K1K2K1+K2)2ε0Ad(K1+K2)

the above effective capacitance is series combination of K1ϵ0A2d and K2ϵ0A2d∴1Cbf=dK1ϵ0A2+dK2ϵ0A2⇒Cbf=2ϵ0Ad(K1K2K1+K2)

the above effective capacitance is series combination of $\frac{K_{1} ϵ_{0} A_{2}}{d}$ and $\frac{K_{2} ϵ_{0} A_{2}}{d}$
$∴ \frac{1}{C_{b f}} = \frac{d}{K_{1} ϵ_{0} A_{2}} + \frac{d}{K_{2} ϵ_{0} A_{2}}$
$\Rightarrow C_{b f} = \frac{2 ϵ_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$