Answer
64.8k+ views
Hint: Understand the structure of the given capacitor carefully and then divide it into a combination of its equivalent series or parallel connection of multiple capacitors. Then use the following mentioned formulas to get the required answer.
Formula Used:
Capacitance of a parallel plate capacitor,$C = \dfrac{{{\varepsilon _ \circ }A}}{d}$
Capacitance of a parallel plate capacitor with dielectric, $C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
In series connection, equivalent capacitance is:
$\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
In parallel connection, equivalent capacitance is:
$C = {C_1} + {C_2}$
Complete step by step answer:
A capacitor is a setup or arrangement of two conductors separated by an insulating medium that is used to store electric charge and electric energy. A capacitor usually consists of two conductors of any size and shape carrying different potentials and charges, and placed closed together in some definite positions relative to one another.
The capacitance of a capacitor is a constant that may be defined as the charge required to be supplied to either of the conductors of the capacitor so as to increase the potential difference between them by a unit amount. Its SI unit is the farad (F).
For a parallel plate capacitor, its capacitance is given as:
$C = \dfrac{{{\varepsilon _ \circ }A}}{d}$
The capacitance of a parallel capacitor when filled with a dielectric of dielectric constant k,
$C = \dfrac{{k{\varepsilon _ \circ }A}}{d} \cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Considering the given setup, we can divide it as two capacitors of capacitance ${C_1}{\text{ and }}{C_2}$ in series as shown below:
![](https://www.vedantu.com/question-sets/331c6e66-95f4-4dc1-bbc3-2342ca8c07b84393398629545356625.png)
So the resultant capacitance of this setup will be C, which will be given by:
$\eqalign{
& \dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \cr
& \Rightarrow \dfrac{d}{{A{\varepsilon _ \circ }k}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \cr} $
Now, the capacitor ${C_1}$has three capacitors inside it which are in parallel connection with each other. So their total capacitance can be calculated using the direct sum of capacitance of each capacitor.
$\eqalign{
& \Rightarrow \dfrac{d}{{A{\varepsilon _ \circ }k}} = \dfrac{1}{{\dfrac{A}{3}{\varepsilon _ \circ }\dfrac{{\left( {{k_1} + {k_2} + {k_3}} \right)}}{{d/2}}}} + \dfrac{1}{{\dfrac{{A{\varepsilon _ \circ }{k_4}}}{{d/2}}}} \cr
& \Rightarrow \dfrac{d}{k} = \dfrac{{3d}}{{2\left( {{k_1} + {k_2} + {k_3}} \right)}} + \dfrac{d}{{2{k_4}}} \cr
& \therefore \dfrac{2}{k} = \dfrac{3}{{\left( {{k_1} + {k_2} + {k_3}} \right)}} + \dfrac{1}{{{k_4}}} \cr} $
Therefore, the correct option is D. i.e., $\dfrac{2}{k} = \dfrac{3}{{{k_1} + {k_2} + {k_3}}} + \dfrac{1}{{{k_4}}}$
Note: In series connection of capacitors, each capacitor has an equal amount of charge for any value of given capacitance. On the other hand in parallel connection of capacitors, the potential differences of the capacitors connected in parallel are equal for any value of capacitances.
Formula Used:
Capacitance of a parallel plate capacitor,$C = \dfrac{{{\varepsilon _ \circ }A}}{d}$
Capacitance of a parallel plate capacitor with dielectric, $C = \dfrac{{k{\varepsilon _ \circ }A}}{d}$
In series connection, equivalent capacitance is:
$\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$
In parallel connection, equivalent capacitance is:
$C = {C_1} + {C_2}$
Complete step by step answer:
A capacitor is a setup or arrangement of two conductors separated by an insulating medium that is used to store electric charge and electric energy. A capacitor usually consists of two conductors of any size and shape carrying different potentials and charges, and placed closed together in some definite positions relative to one another.
The capacitance of a capacitor is a constant that may be defined as the charge required to be supplied to either of the conductors of the capacitor so as to increase the potential difference between them by a unit amount. Its SI unit is the farad (F).
For a parallel plate capacitor, its capacitance is given as:
$C = \dfrac{{{\varepsilon _ \circ }A}}{d}$
The capacitance of a parallel capacitor when filled with a dielectric of dielectric constant k,
$C = \dfrac{{k{\varepsilon _ \circ }A}}{d} \cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Considering the given setup, we can divide it as two capacitors of capacitance ${C_1}{\text{ and }}{C_2}$ in series as shown below:
![](https://www.vedantu.com/question-sets/331c6e66-95f4-4dc1-bbc3-2342ca8c07b84393398629545356625.png)
So the resultant capacitance of this setup will be C, which will be given by:
$\eqalign{
& \dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \cr
& \Rightarrow \dfrac{d}{{A{\varepsilon _ \circ }k}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} \cr} $
Now, the capacitor ${C_1}$has three capacitors inside it which are in parallel connection with each other. So their total capacitance can be calculated using the direct sum of capacitance of each capacitor.
$\eqalign{
& \Rightarrow \dfrac{d}{{A{\varepsilon _ \circ }k}} = \dfrac{1}{{\dfrac{A}{3}{\varepsilon _ \circ }\dfrac{{\left( {{k_1} + {k_2} + {k_3}} \right)}}{{d/2}}}} + \dfrac{1}{{\dfrac{{A{\varepsilon _ \circ }{k_4}}}{{d/2}}}} \cr
& \Rightarrow \dfrac{d}{k} = \dfrac{{3d}}{{2\left( {{k_1} + {k_2} + {k_3}} \right)}} + \dfrac{d}{{2{k_4}}} \cr
& \therefore \dfrac{2}{k} = \dfrac{3}{{\left( {{k_1} + {k_2} + {k_3}} \right)}} + \dfrac{1}{{{k_4}}} \cr} $
Therefore, the correct option is D. i.e., $\dfrac{2}{k} = \dfrac{3}{{{k_1} + {k_2} + {k_3}}} + \dfrac{1}{{{k_4}}}$
Note: In series connection of capacitors, each capacitor has an equal amount of charge for any value of given capacitance. On the other hand in parallel connection of capacitors, the potential differences of the capacitors connected in parallel are equal for any value of capacitances.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)