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Question: A parallel plate capacitor with air as the medium between the plates has a capacitance of\(10\mu F\)...

A parallel plate capacitor with air as the medium between the plates has a capacitance of10μF10\mu F . The area of the capacitor is divided into two equal halves and filled with two media having dielectric constant k1=2{{k}_{1}}=2 andk2=4{{k}_{2}}=4. The capacitance will now be
A. 10μF B. 20μF C. 30μF D. 40μF \begin{aligned} & \text{A}\text{. }10\mu F \\\ & \text{B}\text{. }20\mu F \\\ & \text{C}\text{. }30\mu F \\\ & \text{D}\text{. }40\mu F \\\ \end{aligned}

Explanation

Solution

As the initial capacitor is divided by two equal parts by area and filled with a medium having dielectric constant k1=2{{k}_{1}}=2 andk2=4{{k}_{2}}=4, so the initial capacitor is replaced by two capacitors connected in parallel having same distance between the plates but area of cross-section of the plate is halved. Use the formula for a combination of capacitors in parallel to get the capacitance when the dielectrics are introduced inside the plates.

Formula used:
Capacitance of parallel plate capacitor is given by
C=ϵAdC=\dfrac{\epsilon A}{d}, Where
ϵ= permittivity of medium inside the plates of capacitor A=Area of the plates d=distance between the plates \begin{aligned} & \epsilon =\text{ permittivity of medium inside the plates of capacitor} \\\ & A=\text{Area of the plates} \\\ & d=\text{distance between the plates} \\\ \end{aligned}
Equivalent capacitance of two capacitor having capacitance C1{{C}_{1}}and C2{{C}_{2}}connected in parallel is
C=C1+C2C={{C}_{1}}+{{C}_{2}}
Dielectric constant k of any medium is defined as the ratio of its permittivity ϵ\epsilon to the permittivity of vacuum ϵ0\epsilon _0, i.e.
k=ϵϵ0k=\dfrac{\epsilon }{{{\epsilon }_{0}}}
So permittivity of any medium is ϵ=kϵ0\epsilon =k{{\epsilon }_{0}}

Complete step by step answer:
Capacitance of a parallel plate capacitor having plate area A and distance between the plate d and permittivity of medium between the plate ϵ\epsilon is given by
C=ϵAdC=\dfrac{\epsilon A}{d}
If the medium between the plate is air then ϵ=ϵ0\epsilon ={{\epsilon }_{0}}
So C=ϵ0AdC=\dfrac{\epsilon {{}_{0}}A}{d}
Given C=10μFC=10\mu F.
Now the area of the capacitor is divided into two equal halves and filled with two media having dielectric constant k1=2{{k}_{1}}=2 andk2=4{{k}_{2}}=4.
So now the initial capacitor is now a combination of two parallel plate capacitors connected in parallel having plate areaA2\dfrac{A}{2}.
Let the upper half is filled with a dielectric medium having dielectric constant k1=2{{k}_{1}}=2 and the lower half is filled with a dielectric medium having dielectric constant k2=4{{k}_{2}}=4.
The upper half is now a parallel plate capacitor with plate area A1{{A}_{1}} and distance between the plates is ddand having a dielectric medium in between them with dielectric constant k1{{k}_{1}}.
Its capacitance is given by
C1=ϵ1A1d=k1ϵ0A1d{{C}_{1}}=\dfrac{{{\epsilon }_{1}}{{A}_{1}}}{d}=\dfrac{{{k}_{1}}{{\epsilon }_{0}}{{A}_{1}}}{d} (ϵ1=k1ϵ0)\left( \because {{\epsilon }_{1}}={{k}_{1}}\epsilon {{}_{0}} \right)
Similarly the lower half is now a parallel plate capacitor with plate area A2{{A}_{2}} and distance between the plates is dd and having dielectric medium in between them with dielectric constant k2{{k}_{2}}.
Its capacitance is given by
C2=ϵ2A2d=k2ϵ0A2d{{C}_{2}}=\dfrac{\epsilon {{}_{2}}{{A}_{2}}}{d}=\dfrac{{{k}_{2}}{{\epsilon }_{0}}{{A}_{2}}}{d} (ϵ2=k2ϵ0)\left( \because {{\epsilon }_{2}}={{k}_{2}}\epsilon {{}_{0}} \right)
Equivalent capacitance of parallel combination of C1 and C2{{C}_{1}}\text{ and }{{C}_{2}} is given by
Cp=C1+C2=k1ϵ0A1d+k2ϵ0A2d=ϵ0d(k1A1+k2A2){{C}_{p}}={{C}_{1}}+{{C}_{2}}=\dfrac{{{k}_{1}}{{\epsilon }_{0}}{{A}_{1}}}{d}+\dfrac{{{k}_{2}}{{\epsilon }_{0}}{{A}_{2}}}{d}=\dfrac{{{\epsilon }_{0}}}{d}\left( {{k}_{1}}{{A}_{1}}+{{k}_{2}}{{A}_{2}} \right)
Put the value of k1=2,k2=4,A1=A2=A2{{k}_{1}}=2,{{k}_{2}}=4,{{A}_{1}}={{A}_{2}}=\dfrac{A}{2}
=ϵ0d[2×(A2)+4×(A2)]=ϵ0Ad(1+2) =3ϵ0Ad=3C \begin{aligned} & =\dfrac{{{\epsilon }_{0}}}{d}\left[ 2\times \left( \dfrac{A}{2} \right)+4\times \left( \dfrac{A}{2} \right) \right]=\dfrac{{{\epsilon }_{0}}A}{d}\left( 1+2 \right) \\\ & =\dfrac{3{{\epsilon }_{0}}A}{d}=3C \\\ \end{aligned}
Where
C=ϵ0AdC=\dfrac{\epsilon {{}_{0}}A}{d} is capacitor of initial plate

Given C=10μFC=10\mu F
So
Cp=3×C=3×10μF=30μF{{C}_{p}}=3\times C=3\times 10\mu F=30\mu F

So the correct option is C. 30μF30\mu F

Additional Information
Capacitance is defined as the ratio of the change of electric charge of a system to the change in electric potential. i.e.
C=ΔqΔVC=\dfrac{\Delta q}{\Delta V}
The unit of capacitance is Farad written as F.
The capacitance of a capacitor is one Farad when one coulomb of charge changes the potential between the plates by one Volt.
Capacitance is also defined as the ability of a circuit to collect and store energy.

Note:
In a problem like this 1st see whether all the units are given in S.I or not. If not S.I then convert to S.I and then calculate.
Also note that the capacitance value always expressed in micro Farad or nano Farad or picoFarad, because its value in S.I units is very small.1μF=106F , 1nF=109F and 1pF=1012F1\mu F={{10}^{-6}}F\text{ , }1nF={{10}^{-9}}F\text{ and 1pF=1}{{\text{0}}^{-12}}F
S.I base unit of Farad is F = A2 s4 kg1 m2F\text{ }=\text{ }{{A}^{2}}~{{s}^{4}}~k{{g}^{-1}}~{{m}^{-2}}
A=ampere and s=second,
Dimensional formula [F]=[M1 L2 T4 A2]\left[ F \right]=\left[ {{M}^{-}}^{1}~{{L}^{-}}^{2}~{{T}^{4}}~{{A}^{2}} \right]