Question

Two dielectric slabs of dielectric constants K, and K, have been inserted in between the plates of a capacitor
as shown below. What will be the capacitance of the capacitor inserted ?
(Plate area A)
K,
d/2
---
K,
d/2​

Answer 1

capacitance with dielectrics has been introduced capacitors

εoA ( K1 + K2 ) / d

Explanation:

C= k εoA/d

C = C1 + C2

= εo (A/2) K¹/ d + εo (A/2) K²/ d

= εoA (K1 + K2 ) / d

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