Question

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

Answer 1

Hint: As the initial capacitor is divided by two equal parts by area and filled with a medium having dielectric constant ${{k}_{1}}=2$ and${{k}_{2}}=4$, 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
$C=\dfrac{\epsilon A}{d}$, Where
 $\begin{align}
  & \epsilon =\text{ permittivity of medium inside the plates of capacitor} \\
 & A=\text{Area of the plates} \\
 & d=\text{distance between the plates} \\
\end{align}$
Equivalent capacitance of two capacitor having capacitance ${{C}_{1}}$and ${{C}_{2}}$connected in parallel is
$C={{C}_{1}}+{{C}_{2}}$
Dielectric constant k of any medium is defined as the ratio of its permittivity $\epsilon$ to the permittivity of vacuum $\epsilon _0$, i.e.
$k=\dfrac{\epsilon }{{{\epsilon }_{0}}}$
So permittivity of any medium is $\epsilon =k{{\epsilon }_{0}}$

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 $\epsilon$ is given by
$C=\dfrac{\epsilon A}{d}$
If the medium between the plate is air then \[\epsilon ={{\epsilon }_{0}}\]
So $C=\dfrac{\epsilon {{}_{0}}A}{d}$
Given $C=10\mu F$.
Now the area of the capacitor is divided into two equal halves and filled with two media having dielectric constant ${{k}_{1}}=2$ and${{k}_{2}}=4$.
So now the initial capacitor is now a combination of two parallel plate capacitors connected in parallel having plate area$\dfrac{A}{2}$.
Let the upper half is filled with a dielectric medium having dielectric constant ${{k}_{1}}=2$ and the lower half is filled with a dielectric medium having dielectric constant ${{k}_{2}}=4$.
The upper half is now a parallel plate capacitor with plate area ${{A}_{1}}$ and distance between the plates is $d$and having a dielectric medium in between them with dielectric constant ${{k}_{1}}$.
Its capacitance is given by
${{C}_{1}}=\dfrac{{{\epsilon }_{1}}{{A}_{1}}}{d}=\dfrac{{{k}_{1}}{{\epsilon }_{0}}{{A}_{1}}}{d}$ $\left( \because {{\epsilon }_{1}}={{k}_{1}}\epsilon {{}_{0}} \right)$
Similarly the lower half is now a parallel plate capacitor with plate area ${{A}_{2}}$ and distance between the plates is $d$ and having dielectric medium in between them with dielectric constant ${{k}_{2}}$.
Its capacitance is given by
\[{{C}_{2}}=\dfrac{\epsilon {{}_{2}}{{A}_{2}}}{d}=\dfrac{{{k}_{2}}{{\epsilon }_{0}}{{A}_{2}}}{d}\] \[\left( \because {{\epsilon }_{2}}={{k}_{2}}\epsilon {{}_{0}} \right)\]
Equivalent capacitance of parallel combination of ${{C}_{1}}\text{ and }{{C}_{2}}$ is given by
${{C}_{p}}={{C}_{1}}+{{C}_{2}}=\dfrac{{{k}_{1}}{{\epsilon }_{0}}{{A}_{1}}}{d}+\dfrac{{{k}_{2}}{{\epsilon }_{0}}{{A}_{2}}}{d}=\dfrac{{{\epsilon }_{0}}}{d}\left( {{k}_{1}}{{A}_{1}}+{{k}_{2}}{{A}_{2}} \right)$
Put the value of ${{k}_{1}}=2,{{k}_{2}}=4,{{A}_{1}}={{A}_{2}}=\dfrac{A}{2}$
$\begin{align}
  & =\dfrac{{{\epsilon }_{0}}}{d}\left[ 2\times \left( \dfrac{A}{2} \right)+4\times \left( \dfrac{A}{2} \right) \right]=\dfrac{{{\epsilon }_{0}}A}{d}\left( 1+2 \right) \\
 & =\dfrac{3{{\epsilon }_{0}}A}{d}=3C \\
\end{align}$
Where
$C=\dfrac{\epsilon {{}_{0}}A}{d}$ is capacitor of initial plate

Given $C=10\mu F$
So
${{C}_{p}}=3\times C=3\times 10\mu F=30\mu F$

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

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.
$C=\dfrac{\Delta q}{\Delta V}$
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.$1\mu F={{10}^{-6}}F\text{ , }1nF={{10}^{-9}}F\text{ and 1pF=1}{{\text{0}}^{-12}}F$
S.I base unit of Farad is \[F\text{ }=\text{ }{{A}^{2}}~{{s}^{4}}~k{{g}^{-1}}~{{m}^{-2}}\]
A=ampere and s=second,
Dimensional formula $\left[ F \right]=\left[ {{M}^{-}}^{1}~{{L}^{-}}^{2}~{{T}^{4}}~{{A}^{2}} \right]$