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?

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A08
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#

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