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    the capacity of a parallel plate condenser with air as dielectric is 2

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    The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0. If the air is replaced by glass dielectric constant =5 between the plates, the capacity of the plates and the energy stored in it will respectively be

    The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively C0 and W0. If the air is replaced by glass dielectric constant =5 between the plates, the capacity of the plates and the energy stored in it will respectively be

    Byju's Answer Standard XII Biology

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    The capacity ... Question

    The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively

    C 0 and W 0

    . If the air is replaced by glass (dielectric constant

    = 5

    ) between the plates, the capacity of the plates and the energy stored in it will respectively be

    A C 0 5 , 5 W 0 B C 0 5 , W 0 5 C 5 C 0 , 5 W 0 D 5 C 0 , W 0 5 Open in App Solution

    The correct option is D

    5 C 0 , W 0 5

    We know that after inserting dielectric capacitance of system becomes dielectric constant times original capacitance .

    capacitance with dielectric

    C ′ = ϵ 0 k A d = k C 0 Hence, C ′ = 5 C 0 Also, initial energy W 0 = q 2 2 C 0

    After inserting dielectric, charge will remain the same ( assuming capacitor is isolated, so charge is conserved )

    Thus, Final energy = q 2 2 C ′ = q 2 ( 2 ) ( 5 C 0 ) = W 0 / 5

    Thus, capacitance and final energy stored is

    5 C 0 and W 0 5 respectively.

    Hence, correct option is (B).

    Suggest Corrections 0 SIMILAR QUESTIONS

    Q. A parallel plate condenser initially has air as a medium between the plates. If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity will be ___Q. The energy and capacity of a charged parallel plate capacitor are

    E and C

    respectively. Now a dielectric slab of

    ϵ r = 6

    is inserted in it, then energy and capacity becomes

    (Assuming charge on plates remains constant)

    Q. The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively

    C 0 and W 0

    . If the air is replaced by glass (dielectric constant

    = 5

    ) between the plates, the capacity of the plates and the energy stored in it will respectively be

    Q.

    in a parallel plate condenser the radius of each circular plate is 10cm and the distance between the plates is 5mm. there is a glass slab of 3mm thick and of radius 12 cm with dielectric constant 6 b/w its plates. the capacity of the condenser will be

    Q. When dielectric medium of constant k is filled between the plates of a charged parallel-plate condenser, then the energy stored becomes, as compared to its previous value,

    View More

    स्रोत : byjus.com

    The capacity of a parallel plate condenser with air as a dielectric is 2 mu F . The space between the plates is filled with a dielectric slab with K = 5 . It is charged to a potential of 200V and disconnected from a cell. Work was done in removing the slab from the condenser completely

    Click here👆to get an answer to your question ✍️ The capacity of a parallel plate condenser with air as a dielectric is 2 mu F . The space between the plates is filled with a dielectric slab with K = 5 . It is charged to a potential of 200V and disconnected from a cell. Work was done in removing the slab from the condenser completely

    Question

    The capacity of a parallel plate condenser with air as a dielectric is 2μF. The space between the plates is filled with a dielectric slab with K=5. It is charged to a potential of 200V and disconnected from a cell. Work was done in removing the slab from the condenser completely

    A

    0.8J

    B

    0.6J

    C

    1.2J

    D

    0.032J

    Medium Open in App Solution Verified by Toppr

    Correct option is D)

    Given,

    Capacity, C=2μF=2×10

    −6 F Voltage, V=200V

    Dielectric constant, K=5

    We have, Work done, W= 2 1 ​ CV 2 (1− K 1 ​ ) W= 2 1 ​ ×2×10 −6 ×200 2 (1− 5 1 ​ )=0.032J

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    स्रोत : www.toppr.com

    [Solved] A parallel plate condenser has a capacitance 50μF in air

    CONCEPT: The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance. The capacitance of a conduc

    Home Physics Capacitance

    Question

    Download Solution PDF

    A parallel plate condenser has a capacitance 50μF in air and 110μF when immersed in an oil. The dielectric constant ′k′ of the oil is

    0.45 0.55 0.75 2.20

    Answer (Detailed Solution Below)

    Option 4 : 2.20

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    SSC Scientific Assistant Physics 22 Nov 2017 Official Paper (Shift 1)

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    200 Questions 200 Marks 120 Mins

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    Detailed Solution

    Download Solution PDF

    CONCEPT:

    The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance.

    The capacitance of a conductor is the ratio of charge (Q) given to the conduction to the rise in its potential (V).

    ​C = Q/V

    The unit of capacitance is the farad, (symbol F ).

    If a dielectric medium is filled in a capacitor then the capacitance of the capacitor changes.

    The capacitance, C = K C0 where C0 is the capacitance without the dielectric.

    So Dielectric constant (k) = C/C0

    The dielectric constant of a medium is the ratio of the electrostatic force of interaction between two given point charges held certain distance apart in vacuum/air to the force of interaction between the same two charges held the same distance apart in the material medium.

    CALCULATION:

    Given that:

    Capacitance in air (C0) = 50 μF

    Capacitance in oil (C) = 110 μF

    ⇒K=CmediumCair=11050=2.20

    Download Solution PDF

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    Q1. Two condensers of capacity C and

    14

    C are connected to a battery of V volt, as shown in figure- Then the work done in charging fully both the condensers is

    Q2. Two identical capacitors C1 and C2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C1 using battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy ?Q3. Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is :Q4. The equivalent capacitance between points A and B in the below-shown figure will be ________ μF.Q5. A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is:Q6. An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle 'α' with the plates. It leaves the plates at angle 'β' with kinetic energy K2. Then the ratio of kinetic energies K1 ∶ K2 will be:Q7. Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be:Q8. The equivalent capacitance of the combination shown in the figure is : दिए गये संयोजन में तुल्य धारिता है :Q9. The capacitance of a parallel plate capacitor with air as medium is 6 µF. With the introduction of a dielectric medium, the capacitance becomes 30 µF. The permittivity of the medium is : (ϵ0 = 8.85 × 10−12 C2 N−1 m−2)Q10. A parallel plate capacitor has a uniform electric field '

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