A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is $A m^{2}$ and the separation is $d$ metre. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. Its capacitance (in farad) will be :

$\frac{2 \in_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 \in_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
$\frac{\in_{0} A K_{1} K_{2}}{2 (d_{2} K_{1} + d_{1} K_{2})}$
$\frac{\in_{0} A}{d} (\frac{K_{1} + K_{2}}{2 K_{1} K_{2}})$

Question

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is $A m^{2}$ and the separation is $d$ metre. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. Its capacitance (in farad) will be :

$\frac{2 \in_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
$\frac{2 \in_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
$\frac{\in_{0} A K_{1} K_{2}}{2 (d_{2} K_{1} + d_{1} K_{2})}$
$\frac{\in_{0} A}{d} (\frac{K_{1} + K_{2}}{2 K_{1} K_{2}})$

Toppr Admin · Accepted Answer

Capacitance of first half of the capacitor of width d-2-C1-K1A-x3F5-0d-2-2K1A-x3F5-0dSimilarly the capacitance of other half -C2-2K2A-x3F5-0dThese two capacitance are connected in series-Thus- the net capacitance of the combination is- Ceff-C1C2C1-C2-2A-x3F5-0d-K1K2K1-K2-

A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is Am2 and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance (in farad) will be :2∈0Ad(K1+K2K1K2)2∈0Ad(K1K2K1+K2)∈0AK1K22(d2K1+d1K2)∈0Ad(K1+K22K1K2)

Capacitance of first half of the capacitor of width d/2=C1=K1Aϵ0d/2=2K1Aϵ0dSimilarly the capacitance of other half =C2=2K2Aϵ0dThese two capacitance are connected in series.Thus, the net capacitance of the combination is, Ceff=C1C2C1+C2=2Aϵ0d(K1K2K1+K2)