Questions & Answers
Question
AnswersThe 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.
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.