If the area of plates is tripled and medium of relative permitivity 2 is introduced between plates then capacitance of capacitor isa)increases by 500 percent b)decreases by 500 percent c) increases by 600 percent?

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Answer 1 · 2018-01-10T08:24:10

The answer is option #(a)" increases by " 500 %#

#C=(kxxepsilon_0xxA)/d#

Where

Area of the plates #=A#

Separation between the plates #=d#

The permittivity of space is #=epsilon_0=8.854*10^-12Fm^-1#

The relative permittivity of the dielectric material between the plates

#A_1=3A#

#k_1=2k#

Therefore,

#C_1=(k_1xxepsilon_0xxA_1)/d=(2kxxepsilon_0xx3A)/d=6(kxxepsilon_0xxA)/d#

#C_1=6C#

The increase in the capacitance is

#C_1-C=6C-C=5C#

The increase is #=500%#