# The capacitance of a parallel plate capacitor with air as a medium is $3\mu F$ . With the introduction of a dielectric medium between the plates, the capacitance becomes $15\mu F$ . The permittivity of the medium is:(A) $5{C^2}{N^{ - 1}}{m^{ - 2}}$ (B) $10{C^2}{N^{ - 1}}{m^{ - 2}}$ (C) $0.44 \times {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}$ (D) $8.854 \times {10^{ - 11}}{C^2}{N^{ - 1}}{m^{ - 2}}$

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Hint: To solve this question using the formula of capacitance between two parallel plates i.e. $c = \dfrac{{A{\varepsilon _0}}}{d}$ and when any dielectric is inserted between the capacitor plates, the capacitance increases by a factor i.e. new capacitance ${c_d} = \dfrac{{kA{\varepsilon _0}}}{d}$ . Now, compare both the equations and use the formula of permittivity that is $\varepsilon = {\varepsilon _0}k$ and you will get the desired result.

Complete step by step solution
Let us take the original capacitance to be $\prime c\prime$ and the new capacitance, when the dielectric is inserted between the plates, be ${c_d}^{}$ .
Using the formula of capacitance, $c = \dfrac{{A{\varepsilon _0}}}{d}$ ....... $\left( 1 \right)$
When a dielectric is inserted between plates, the capacitance of capacitor increases by the factor of $k$ (dielectric constant) i.e. the new formula is ${c_d} = \dfrac{{kA{\varepsilon _0}}}{d}$ ............ $\left( 2 \right)$
where $A$ is the area of the capacitor is plates and $d$ is the distance between the two parallel plates of the capacitor.
Now, as per the given question, the original capacitance $c = 3\mu F$ .......... $\left( 3 \right)$
And, capacitance, when the dielectric is inserted, is ${c_d} = 15\mu F$ ......... $\left( 4 \right)$
Dividing equation $\left( 1 \right)$ and $\left( 2 \right)$ , we get:
$\dfrac{c}{{{c_d}}} = \dfrac{{\dfrac{{A{\varepsilon _0}}}{d}}}{{\dfrac{{kA{\varepsilon _0}}}{d}}}$ ............. $\left( 5 \right)$
$\dfrac{c}{{{c_d}}} = \dfrac{1}{k}$ ........... $\left( 6 \right)$
Now, putting the values from the equation $\left( 3 \right)$ and $\left( 4 \right)$ in the equation $\left( 6 \right)$ :
We get, $\dfrac{{3\mu F}}{{15\mu F}} = \dfrac{1}{k}$
Solving this, we get $k = 5$
Now, to calculate the permittivity we will sue the formula $\varepsilon = {\varepsilon _0}k$
where the value of ${\varepsilon _0}$ is permittivity constant and it is ${\varepsilon _0} = 8.85 \times {10^{ - 12}}$
When a dielectric is inserted in between the capacitor plates, the permittivity constant also increased by a factor of $k$ , so the new permittivity will be
$\varepsilon = {\varepsilon _0}k$
Putting the value of $k$ and ${\varepsilon _0}$ , it becomes
$= 8.85 \times {10^{ - 12}} \times 5$
$\Rightarrow 0.44 \times {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}$
Therefore, the answer is an option (C) $0.44 \times {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}$ .

Note
Remember in such questions it is very obvious to make unit mistakes. So, always take care of units and ${\varepsilon _0}$ is a constant, its value is fixed in every question and dielectric constant is different for every material, don’t get confused with the word constant.