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    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick

    Dec 24,2022 - In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? | EduRev Physics Question is disucussed on EduRev Study Group by 5048 Physics Students.

    Physics Question  >  In a parallel plate capacitor the distance be...

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is

    a) b) c) d)

    Correct answer is option 'A'. Can you explain this answer?

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    SUNITA BISHNOI Sep 10, 2019 Related

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?

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    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? for Physics 2022 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Physics 2022 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?.

    Solutions for In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.

    Here you can find the meaning of In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thick- ness 5 cm each and dielectric constants K1= 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Physics tests.

    स्रोत : edurev.in

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs o thickness 5 cm each and dielectric constants K1 = 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is:

    Click here👆to get an answer to your question ✍️ In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs o thickness 5 cm each and dielectric constants K1 = 2 and K2 = 4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is: -

    In a parallel plate capacitor the distance between the plates is 10cm. Two dielectric slabs o thickness 5cm each and dielectric constants K

    Question 1 ​ =2 and K 2 ​

    =4 respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is:-

    A

    3 2000 ​ ϵ 0 ​

    B

    3 1000 ​ ϵ 0 ​

    C−250ϵ

    0 ​

    D

    3 2000 ​ ϵ 0 ​ Medium Open in App

    स्रोत : www.toppr.com

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric constants ${K_1} = 2$ and ${K_2} = 4$ respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is:A. $

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric constants ${K_1} = 2$ and ${K_2} = 4$ respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. Th...

    In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric constants

    K 1 =2 K1=2 and K 2 =4 K2=4

    respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is:

    A. − 2000 3 ε ∘ −20003ε∘ B. − 1000 3 ε ∘ −10003ε∘ C. −250 ε ∘ −250ε∘ D. 2000 3 ε ∘ 20003ε∘ Answer Verified 225.9k+ views

    Hint: Surface charge density is defined as the measure of amount of electric charge is accumulated over a surface. In other words, it is a quantity of charge measured per unit surface area. If ‘q’ is the charge and ‘A’ is the area of the surface, then the surface charge density is written as,

    σ=qA σ=qA

    . The SI unit of the surface charge density is expressed in

    C/ m 2 C/m2

    Formula Used:

    V=Ed V=Ed

    Complete answer:

    A parallel plate capacitor is set up by an arrangement of electrodes and a dielectric or an insulating material. It can only store a finite amount of energy before any dielectric breakdown.

    When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, this setup is known as the parallel plate capacitor. In the given parallel plate capacitor with the given distance between the plates the charge density can be given as,

    σ= Q A σ=QA . V = E 1 d+ E 2 d= σ ε 1 d+ σ ε 2 d= σ 2 ε ∘ d+ σ 4 ε ∘ d= 3σ 4 ε ∘ d

    E1d+E2d=σε1d+σε2d=σ2ε∘d+σ4ε∘d=3σ4ε∘d

    It is given that the V = 100 Volts, d =

    5× 10 −2 cm 5×10−2cm ⇒σ= 4 ε ∘ 3d V= 4 ε ∘ 3×5× 10 −2 ×100= 4× 10 4 15 ε ∘

    ⇒σ=4ε∘3dV=4ε∘3×5×10−2×100=4×10415ε∘

    P 1 ¯ ¯ ¯ ¯ ¯ = ε ∘ x E 1 − → = ε ∘ ( K 2 −1) E 1 − →

    P1¯=ε∘xE1→=ε∘(K2−1)E1→

    ⇒ σ 1 = ε ∘ × σ 2 ε ∘ = σ 2 ⇒σ1=ε∘×σ2ε∘=σ2 P ¯ ¯ ¯ ¯ 2 = ε ∘ x E 2 − → = ε ∘ ( K 2 −1) E 2 − →

    P¯2=ε∘xE2→=ε∘(K2−1)E2→

    ⇒ σ 1 =3 ε ∘ × σ 4 ε ∘ = 3σ 4 ⇒σ1=3ε∘×σ4ε∘=3σ4 Now, σ= σ 1 − σ 2 = σ 2 − 3σ 4 =− σ 4 σ=σ1−σ2=σ2−3σ4=−σ4 ⇒σ=− 1 4 × 4× 10 4 15 ε ∘ =− 2000 3 ε ∘

    ⇒σ=−14×4×10415ε∘=−20003ε∘

    Thus, the value of the net bound surface charge density at the interface of the two dielectrics is

    − 2000 3 ε ∘ −20003ε∘ .

    Hence, option (A) is the correct answer.Note:

    The electric potential is defined as the amount of work done in order to move a unit charge from a reference point to a specific point against an electric field. When two plates consist of vacuum between each other, then the potential energy across the capacitor is given by,

    V=Ed V=Ed

    ; where ‘E’ is denoted as a uniform electric field and ‘d’ is expressed as the distance between the two-point charges.

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