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
Let the parallel plate capacitor be filled with two blocks of identical size and dielectric constants
As such area of plates of each dielectric materiel is
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Capacitance
C =( k ε_0 A) / (d)
whereA is area of plates,d separation between the plates,k relative permittivity of the dielectric material between the plates andepsilon_0=8.854xx10^-12\ Fm^-1 is permittivity of space.
k=1 for free space and~~ 1 for air.
It is given that
The arrangement shown in the figure is equivalent two capacitors in parallel. As such total capacitance
:.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