Answer
Verified
429.6k+ views
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.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Give 10 examples for herbs , shrubs , climbers , creepers
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE