Answer
Verified
396.6k+ views
Hint: The above combination can be considered as three capacitors connected in parallel to each other. One with an area of $\dfrac{A}{2}$ with a dielectric present in it and the other two with the area $\dfrac{A}{4}$ and with no dielectric inside it. The resultant capacitance of these capacitors will give the capacitance of the origin capacitor.
Complete step by step answer:
In a configuration given in the figure above, the capacitor can be broken up into three regions of capacitance. The first and third region has the same surface area, and there is no dielectric present in these regions. The middle region or the second region will have a different capacitance compared to the first and the third region since the surface area is different and there is a dielectric present in that region.
So, the capacitance of a parallel plate capacitor of area A and separated by a distance d is given by,
$c=\dfrac{{{\varepsilon }_{0}}A}{d}$
Where ${{\varepsilon }_{0}}$ is the permittivity of free space.
If a dielectric completely cover the distance d of the capacitor, then the resultant capacitance will be,
$c'=K\dfrac{{{\varepsilon }_{0}}A}{d}$
K is the value of the dielectric constant.
So, the capacitance of the first and the third region is,
${{c}_{1}}={{c}_{3}}=\dfrac{{{\varepsilon }_{0}}\left[ \dfrac{A}{4} \right]}{d}$
$\therefore {{c}_{1}}={{c}_{3}}=\dfrac{{{\varepsilon }_{0}}A}{4d}$ …. Equation (1)
The capacitance of the middle layer which contains a dielectric is,
${{c}_{2}}=K\dfrac{{{\varepsilon }_{0}}\left[ \dfrac{A}{2} \right]}{d}$
$\therefore {{c}_{2}}=\dfrac{K{{\varepsilon }_{0}}A}{2d}$ …. Equation (2)
These regions are connected in such a way that the regions are connected as parallel. So, the resultant capacitance of the three regions is given by,
$C={{c}_{1}}+{{c}_{2}}+{{c}_{3}}$
$\Rightarrow C=\dfrac{{{\varepsilon }_{0}}A}{4d}+\dfrac{K{{\varepsilon }_{0}}A}{2d}+\dfrac{{{\varepsilon }_{0}}A}{4d}$
$\Rightarrow C=\left[ \dfrac{K+1}{2} \right]\dfrac{{{\varepsilon }_{0}}A}{d}$
It is given that the value of K is four and the capacitance of the original capacitor with air as the medium is $10\mu F$ .
$\Rightarrow C=\left[ \dfrac{4+1}{2} \right]\times 10\mu F$
$C=25\mu F$
So, the capacitance of the capacitor with the dielectric present is $C=25\mu F$.
So, the answer to the question is option (D).
Note:
When two parallel plates are connected across a battery, the plates are charged, and the electric field is established. This setup is known as the parallel plate capacitor. The direction of the electric field is defined as the direction in which the positive test charge would flow. Capacitance is the limitation of the body to store electric charge.
When a number of capacitors of variable capacitance are connected in parallel, the resultant capacitance is given by,
${{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+...$
When a number of capacitors of variable capacitance are connected in series, the resultant capacitance is given by,
$\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+...$
Complete step by step answer:
In a configuration given in the figure above, the capacitor can be broken up into three regions of capacitance. The first and third region has the same surface area, and there is no dielectric present in these regions. The middle region or the second region will have a different capacitance compared to the first and the third region since the surface area is different and there is a dielectric present in that region.
So, the capacitance of a parallel plate capacitor of area A and separated by a distance d is given by,
$c=\dfrac{{{\varepsilon }_{0}}A}{d}$
Where ${{\varepsilon }_{0}}$ is the permittivity of free space.
If a dielectric completely cover the distance d of the capacitor, then the resultant capacitance will be,
$c'=K\dfrac{{{\varepsilon }_{0}}A}{d}$
K is the value of the dielectric constant.
So, the capacitance of the first and the third region is,
${{c}_{1}}={{c}_{3}}=\dfrac{{{\varepsilon }_{0}}\left[ \dfrac{A}{4} \right]}{d}$
$\therefore {{c}_{1}}={{c}_{3}}=\dfrac{{{\varepsilon }_{0}}A}{4d}$ …. Equation (1)
The capacitance of the middle layer which contains a dielectric is,
${{c}_{2}}=K\dfrac{{{\varepsilon }_{0}}\left[ \dfrac{A}{2} \right]}{d}$
$\therefore {{c}_{2}}=\dfrac{K{{\varepsilon }_{0}}A}{2d}$ …. Equation (2)
These regions are connected in such a way that the regions are connected as parallel. So, the resultant capacitance of the three regions is given by,
$C={{c}_{1}}+{{c}_{2}}+{{c}_{3}}$
$\Rightarrow C=\dfrac{{{\varepsilon }_{0}}A}{4d}+\dfrac{K{{\varepsilon }_{0}}A}{2d}+\dfrac{{{\varepsilon }_{0}}A}{4d}$
$\Rightarrow C=\left[ \dfrac{K+1}{2} \right]\dfrac{{{\varepsilon }_{0}}A}{d}$
It is given that the value of K is four and the capacitance of the original capacitor with air as the medium is $10\mu F$ .
$\Rightarrow C=\left[ \dfrac{4+1}{2} \right]\times 10\mu F$
$C=25\mu F$
So, the capacitance of the capacitor with the dielectric present is $C=25\mu F$.
So, the answer to the question is option (D).
Note:
When two parallel plates are connected across a battery, the plates are charged, and the electric field is established. This setup is known as the parallel plate capacitor. The direction of the electric field is defined as the direction in which the positive test charge would flow. Capacitance is the limitation of the body to store electric charge.
When a number of capacitors of variable capacitance are connected in parallel, the resultant capacitance is given by,
${{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+...$
When a number of capacitors of variable capacitance are connected in series, the resultant capacitance is given by,
$\dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+...$
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE