The value of capacitance of an capacitor is 8uf.Two blocks of identical size and dielectric constants k1=3.0 and k2=6.0 now fill the space between the plates of this parallel plate capacitor . calculate the new value of capacitance?

1 Answer
Aug 3, 2018

Let the parallel plate capacitor be filled with two blocks of identical size and dielectric constants k_1=3.0 and k_2=6.0 as shown in the figure below.

As such area of plates of each dielectric materiel is A/2 with plate separation d.
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Capacitance C of parallel plate capacitor is given by the expression

C =( k ε_0 A) / (d)
where A is area of plates, d separation between the plates, k relative permittivity of the dielectric material between the plates and epsilon_0=8.854xx10^-12\ Fm^-1 is permittivity of space.
k=1 for free space and ~~ 1 for air.

It is given that C=\ 8muF

The arrangement shown in the figure is equivalent two capacitors in parallel. As such total capacitance C_T is sum of individual capacitances.

:.C_T=C_1+C_2
=>C_T=( k_1 ε_0 (A/2)) / (d)+( k_2 ε_0 (A/2)) / (d)
=>C_T=(ε_0 A)/d( k_1 +k_2 ) / (2)
=>C_T=Cxx( k_1 +k_2 ) / (2)

Inserting given values we get

C_T=8xx( 3.0 +6.0 ) / (2)
C_T=36\ muF