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A parallel plate capacitor is of area 6 cm$$^2$$ and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant $$K_1 = 10$$, $$K_2 = 12$$ and $$K_3 = 14$$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:
$$A' = \frac{A}{3}$$
Equivalent capacitance of the parallel combination: $$C_{\text{eq}} = C_1 + C_2 + C_3$$
$$\frac{K_{\text{eq}}\varepsilon_0 A}{d} = \frac{K_1 \varepsilon_0 (A/3)}{d} + \frac{K_2 \varepsilon_0 (A/3)}{d} + \frac{K_3 \varepsilon_0 (A/3)}{d}$$
$$K_{\text{eq}} = \frac{K_1 + K_2 + K_3}{3}$$
$$K_{\text{eq}} = \frac{10 + 12 + 14}{3} = \frac{36}{3} = 12$$
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