Question

A capacitor stores $60\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of $120\mu C$ flows through the battery. The dielectric constant of the material inserted is:
$\begin{align}
  & A)1 \\
 & B)2 \\
 & C)3 \\
\end{align}$
$D)$none

Answer 1

Hint: Capacitance of a capacitor is directly proportional to the plate area of the capacitor and is inversely proportional to the distance between the plates of the capacitor. When filled with a dielectric medium inside the plates of the capacitor, capacitance of the capacitor is also proportional to the dielectric constant of the medium. Charge stored in a capacitor is equal to the product of capacitance of the capacitor and the voltage applied across the capacitor.
Formula used:
$1)C=\dfrac{{{\varepsilon }_{0}}A}{d}$
$2)C'=\dfrac{k{{\varepsilon }_{0}}A}{d}$
$3)Q=CV$

Complete answer:
A capacitor is a passive component in an electric circuit, which can store electric charge in an electric field. Capacitance of a capacitor can be explained as the effect of the capacitor. The SI unit of capacitance is $farad(F)$. Capacitance of a capacitor is given by
$C=\dfrac{{{\varepsilon }_{0}}A}{d}$
where
$C$ is the capacitance of a capacitor
${{\varepsilon }_{0}}$ is the permittivity of free space
$A$ is the area of plates of the capacitor
$d$ is the distance between the plates of the capacitor
Let this be equation 1.
When we fill the space between the plates of the capacitor with a dielectric medium of dielectric constant $k$, the capacitance of the capacitor will become
$C'=\dfrac{k{{\varepsilon }_{0}}A}{d}$
where
$C'$ is the capacitance of a capacitor filled with dielectric inside
$k$ is the dielectric constant of the medium in between the plates of the capacitor
${{\varepsilon }_{0}}$ is the permittivity of free space
$A$ is the area of plates of the capacitor
$d$ is the distance between the plates of the capacitor
Let this be equation 2.
Clearly, from equation 1 and equation 2,
$C'=kC$
where
$C$ is the capacitance of a capacitor with air in between the plates of the capacitor
$C'$ is the capacitance of the capacitor with a dielectric medium in between the plates of the capacitor
$k$ is the dielectric constant of the dielectric medium
Let this be equation 3.
Coming to our question, we are provided with a capacitor, which stores a charge of $60\mu C$, when connected across a battery. We are also told that when the gap between the plates is filled with a dielectric, a charge of $120\mu C$ flows through the battery. We are required to determine the dielectric constant of the material inserted between the plates of the capacitor.
Capacitance of the capacitor before filling the dielectric medium is given by equation 1 and the capacitance of the capacitor after filling with a dielectric medium is given by equation 2.
Now, if the charge stored in the capacitor when there is no dielectric in between the plates is denoted as $Q$, it is given by
$Q=CV$
where
$C$ is the capacitance of the capacitor before filling with dielectric
$V$ is the voltage applied across the capacitor
Let this be equation 4.
Also, if the charge stored in the capacitor when the capacitor is filled with dielectric is denoted as $Q'$, it is given by
$Q'=C'V$
where
$C'$ is the capacitance of the capacitor after filling the dielectric between the plates of the capacitor
$V$ is the voltage applied across the capacitor
Let this be equation 5.
From equation 3, we know that
$C'=kC$
Substituting this value of in equation 5 and then dividing it by equation 4, we have$\dfrac{Q'}{Q}=\dfrac{C'V}{CV}=\dfrac{kCV}{CV}=k$
Let this be equation 6.
From the question, we know that
$Q=60\mu C$
and
$Q'=60\mu C+120\mu C=180\mu C$
Substituting these values in the above expression, we have
$k=\dfrac{180\mu C}{60\mu C}=3$
Therefore, the dielectric constant of the dielectric medium filled in between the plates of the capacitor is equal to $3$.

Hence, the correct answer is option $C$.

Note:
It is given that the charge stored by the capacitor before it is filled with dielectric is $60\mu C$. Now, when a dielectric of dielectric constant equal to $3$ is filled in between the plates of the capacitor, it is said that an additional charge of $120\mu C$ can also be stored inside the capacitor. This suggests that the total charge which can be stored inside the capacitor with the dielectric is the sum of the already stored charge and the additional charge, which flows from the battery to the capacitor. Students need to realise this is how, the total capacitance in the second case turned out to be $180\mu C$.