
In a parallel plate capacitor set up, the plate area of the capacitor is 2 $m^2$ and the plates are separated by 1 m. If the space between the plates is filled with a dielectric material of thickness 0.5 m and area 2 $m^2$ (see fig) the capacitance of the set-up will be ________$\varepsilon_0$. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer)

Answer
163.5k+ views
Hint: In this problem, the capacitance of a parallel plate capacitor must be determined. $C=\dfrac{K{{\in }_{o}}A}{d}$ is the formula for a parallel plate capacitor. The net capacitance for capacitors connected in parallel is calculated using the formula \[{{C}_{net}}={C_1}+{C_2}\]. Additionally, the charge of a parallel plate capacitor is determined by$Q=CV$, and the parallel plate capacitor's electric field is determined by\[E=Vd\].
Complete answer:
The ratio of the magnitude of the charge on each conductor to the magnitude of the potential difference (PD) between them is known as the capacitance (C) of a capacitor.
Hence, we can say that capacitance is equal to the ratio charge is to potential difference.
The value of capacitance is always positive. The coulomb per volt or farad is the S.I. unit for capacitance (F).
The size, shape, relative placements of the plates, and the medium between them all affect the capacitance value. The charge on the plate or the distance between the plates has no bearing on the value of C.
The capacitor's capacitance is determined by $C=\dfrac{{{\varepsilon }_{0}}A}{d}$.
According to the question,
${{C}_{1}}=\dfrac{K{{\varepsilon }_{0}}A}{{d}/{2}\;}$ and ${{C}_{2}}=\dfrac{{{\varepsilon }_{0}}A}{{d}/{2}\;}$
OR
$C=\dfrac{{{E}_{0}}A}{\varepsilon \left( \dfrac{d}{K} \right)}=\dfrac{{{E}_{0}}A}{\dfrac{d}{2K}+\dfrac{d}{2}}$
Now,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2K{{\varepsilon }_{0}}A}+\dfrac{d}{2{{\varepsilon }_{0}}A}$
$\Rightarrow \dfrac{1}{C}=\dfrac{2{{E}_{0}}AK}{d(K+1)}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2{{\varepsilon }_{0}}A}\left( \dfrac{K+1}{K} \right)$
$C=\dfrac{2{{\varepsilon }_{0}}AK}{d(K+1)}=\dfrac{2\times 2\times 3.2}{1\times 4.2}{{E}_{0}}$
$C=3.04{{E}_{0}}$
Hence, the answer is 3.04.
Note: Sparking occurs when the two capacitor plates are connected by a conducting wire, demonstrating the transformation of electrical energy into heat and light energy. If a medium with a dielectric constant of K is put between a capacitor's plates, the capacitance of the capacitor will increase K times.
Complete answer:
The ratio of the magnitude of the charge on each conductor to the magnitude of the potential difference (PD) between them is known as the capacitance (C) of a capacitor.
Hence, we can say that capacitance is equal to the ratio charge is to potential difference.
The value of capacitance is always positive. The coulomb per volt or farad is the S.I. unit for capacitance (F).
The size, shape, relative placements of the plates, and the medium between them all affect the capacitance value. The charge on the plate or the distance between the plates has no bearing on the value of C.
The capacitor's capacitance is determined by $C=\dfrac{{{\varepsilon }_{0}}A}{d}$.
According to the question,
${{C}_{1}}=\dfrac{K{{\varepsilon }_{0}}A}{{d}/{2}\;}$ and ${{C}_{2}}=\dfrac{{{\varepsilon }_{0}}A}{{d}/{2}\;}$
OR
$C=\dfrac{{{E}_{0}}A}{\varepsilon \left( \dfrac{d}{K} \right)}=\dfrac{{{E}_{0}}A}{\dfrac{d}{2K}+\dfrac{d}{2}}$
Now,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2K{{\varepsilon }_{0}}A}+\dfrac{d}{2{{\varepsilon }_{0}}A}$
$\Rightarrow \dfrac{1}{C}=\dfrac{2{{E}_{0}}AK}{d(K+1)}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2{{\varepsilon }_{0}}A}\left( \dfrac{K+1}{K} \right)$
$C=\dfrac{2{{\varepsilon }_{0}}AK}{d(K+1)}=\dfrac{2\times 2\times 3.2}{1\times 4.2}{{E}_{0}}$
$C=3.04{{E}_{0}}$
Hence, the answer is 3.04.
Note: Sparking occurs when the two capacitor plates are connected by a conducting wire, demonstrating the transformation of electrical energy into heat and light energy. If a medium with a dielectric constant of K is put between a capacitor's plates, the capacitance of the capacitor will increase K times.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Charging and Discharging of Capacitor
