A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k₁ = 3 and thickness d/3 while the other one has dielectric constant k₂ = 6 and thickness 2d/3 . Capacitance of the capacitor is now

1.8 pF

45 pF

40.5 pF

20.25 pF

Solution

40.5 pF

fraction numerator Aε subscript 0 over denominator begin display style straight d subscript 1 over 3 plus straight d subscript 2 over 6 end style end fraction space equals space fraction numerator Aε subscript 0 over denominator begin display style straight d over 9 plus fraction numerator 2 straight d over denominator 18 end fraction end style end fraction space equals space fraction numerator 18 space Aε subscript 0 over denominator 4 straight d end fraction

C' = 40.5 PF

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