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PhysicsNEET 2025Conductors and dielectrics

Electrostatic Potential and Capacitance

The plates of a parallel plate capacitor are separated by dd. Two slabs of different dielectric constants K1K_1 and K2K_2, with thicknesses 38d\dfrac{3}{8}d and d2\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=1.25K2K_1 = 1.25K_2, the value of K1K_1 is:

  1. A.2.66✓
  2. B.2.33
  3. C.1.60
  4. D.1.33

Solution

Effective gap =3d/8K1+d/2K2+d8= \frac{3d/8}{K_1} + \frac{d/2}{K_2} + \frac{d}{8} (air). Capacitance doubling means this equals d2\frac{d}{2}: 38K1+12K2=38\frac{3}{8K_1} + \frac{1}{2K_2} = \frac{3}{8}. With K2=0.8K1K_2 = 0.8K_1: 38K1+58K1=38⇒K1=83≈2.66\frac{3}{8K_1} + \frac{5}{8K_1} = \frac{3}{8} \Rightarrow K_1 = \frac{8}{3} \approx 2.66.

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