The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constants K1 and K2, with thicknesses \dfrac{3}{8}d and \dfrac{d}{2} respectively, are inserted in the ca

Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constants K1 and K2, with thicknesses \dfrac{3}{8}d and \dfrac{d}{2} respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25K2, the value of K1 is:

Vishal Gupta · Accepted Answer

A. Effective gap = \frac{3d/8}{K1} + \frac{d/2}{K2} + \frac{d}{8} (air). Capacitance doubling means this equals \frac{d}{2}: \frac{3}{8K1} + \frac{1}{2K2} = \frac{3}{8}. With K2 = 0.8K1: \frac{3}{8K1} + \frac{5}{8K1} = \frac{3}{8} \Rightarrow K1 = \frac{8}{3} \approx 2.66.

Electrostatic Potential and Capacitance

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