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Two dielectric slabs of dielectric constants K_(1) and K_(2) are filled in between the two plates, eachof area A of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor. |
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Answer» Solution :The ARRANGEMENT is equivalent to a parallel of two capacitors, each with plate area `A/2` and separation d. So net capacitance is `C=C_(1)+C_(2) =(epsi_(0) ""A/2 K_(1))/(d)(epsi_(0)""A/2 K_(2))/(d)` `=(epsi_(0) A(K_(1)+K_(2)))/(2d)` 
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