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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}} $

Answer
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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.