# A Parallel Plate Capacitor Has Capacitance C. If It Is Equally Filled The Parallel Of Materials Of Dielectric Constant K1 And K2 Its Capacity Becomes C1 . The Ratio Of C1 And C Is

$$C_{A} = \frac{K_{1}\epsilon _{0}A}{\frac{d}{2}}$$ $$C_{B} = \frac{K_{2}\epsilon _{0}A}{\frac{d}{2}}$$

Capacitance,

$$C_{eq} = \frac{C_{1}}{C_{2}}$$ $$\Rightarrow \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$$ $$\Rightarrow \frac{C_{A}C_{B}}{C_{A} + C_{B}} = (\frac{2K_{1}K_{2}}{K_{1} + K_{2}}) \frac{\epsilon _{0}A}{d}$$

Since e = $$\frac{\epsilon _{0}A}{d}$$

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