Electrostatic Potential and Capacitance


Electrostatic Potential and Capacitance

  1. Two condensers, one of capacity C and other of capacity C/2 are connected to a V-volt battery, as shown.​​​[2007] ​The work done in charging fully both the condensers is​​​









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    ​Work done = Change in energy ​

    =
    1
    C +
    C
    V² =
    1
    3C
    V² =
    3CV²
    22224

    Correct Option: B

    ​Work done = Change in energy ​

    =
    1
    C +
    C
    V² =
    1
    3C
    V² =
    3CV²
    22224


  1. Two parallel metal plates having charges + Q and –Q face  each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will









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    ​Electric field ​ ​

    E =
    σ
    =
    Q
    ε

    ε of kerosine oil is more than that of air. ​
    As ε increases, E decreases.

    Correct Option: D

    ​Electric field ​ ​

    E =
    σ
    =
    Q
    ε

    ε of kerosine oil is more than that of air. ​
    As ε increases, E decreases.



  1. A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C2 , in terms of C1, is then​​​









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    ​In series, Ceff =
    C1
    n1

    ∴ Energy stored, Es =
    1
    Ceff C2² =
    1
    C1
    16V²
    22n1

    = 8V²
    C1
    n1

    ​In parallel, Ceff = n2 C2 ​ ​
    Energy stored, Ep =
    2
    n2C2
    2

    8V²C1
    =
    1
    = n2C2
    n12

    ⇒ C2 =
    16C1
    n1n2

    Correct Option: D

    ​In series, Ceff =
    C1
    n1

    ∴ Energy stored, Es =
    1
    Ceff C2² =
    1
    C1
    16V²
    22n1

    = 8V²
    C1
    n1

    ​In parallel, Ceff = n2 C2 ​ ​
    Energy stored, Ep =
    2
    n2C2
    2

    8V²C1
    =
    1
    = n2C2
    n12

    ⇒ C2 =
    16C1
    n1n2


  1. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is :









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    ​The energy stored by a capacitor ​​​​​...(i) ​

    U =
    1
    CV²...........(i)
    2

    V is the p.d. between two plates of the capacitor. ​The capacitance of the parallel plate capacitor ​​V = E.d. ​ ​
    C =
    0
    d

    Substituting the value of C in equation (i) ​
    U =
    1
    0
    (Ed)² =
    1
    0E²d
    2d2

    Correct Option: C

    ​The energy stored by a capacitor ​​​​​...(i) ​

    U =
    1
    CV²...........(i)
    2

    V is the p.d. between two plates of the capacitor. ​The capacitance of the parallel plate capacitor ​​V = E.d. ​ ​
    C =
    0
    d

    Substituting the value of C in equation (i) ​
    U =
    1
    0
    (Ed)² =
    1
    0E²d
    2d2



  1. Two thin dielectric slabs of dielectric constants K1 and K2 (K1 ≤ K2) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘E’ between the plates with distance ‘d’ as measured from plate P is correctly shown by :









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    Electric field, E ∝ 1/K ​
    As K1 < K2 so E1 > E2
    Hence graph (c) correctly dipicts the variation of electric field E with distance d.

    Correct Option: C

    Electric field, E ∝ 1/K ​
    As K1 < K2 so E1 > E2
    Hence graph (c) correctly dipicts the variation of electric field E with distance d.