A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each plate is A metre² and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be :

\[\frac{\varepsilon_{0A}}{d}\left( K_{1} +K_{2}\right)\]
\[\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\]
\[\frac{\varepsilon_{0A}}{d}2\left( K_{1} +K_{2}\right)\]
\[\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} -K_{2}\right)}{2}\]
Solution:
Both the parts can be taken as separate capacitors connected in parallel.
So, C=C1+ C2=\(\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\)
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