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NEET-XII-Physics

02: Electrostatic Potential And Capacitance

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    Chapter 2 - Electrostatic Potential And Capacitance
  • Qstn #1
    Two charges 5 × 10-8 C and -3 × 10-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
    Ans : There are two charges,



    Distance between the two charges, d = 16 cm = 0.16 m

    Consider a point P on the line joining the two charges, as shown in the given figure.



    r = Distance of point P from charge q1

    Let the electric potential (V) at point P be zero.

    Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.



    Where,

    = Permittivity of free space

    For V = 0, equation (i) reduces to



    Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

    Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero, as shown in the following figure.



    For this arrangement, potential is given by,



    For V = 0, equation (ii) reduces to



    Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
  • Qstn #2
    A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
    Ans : The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.



    Where,

    Charge, q = 5 µC = 5 × 10-6 C

    Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm

    Distance of each vertex from centre O, d = 10 cm

    Electric potential at point O,



    Where,

    = Permittivity of free space



    Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
  • Qstn #3
    Two charges 2 μC and -2 µC are placed at points A and B 6 cm apart.
  • #3-a
    Identify an equipotential surface of the system.
    Ans : The situation is represented in the given figure.



    An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
  • #3-b
    What is the direction of the electric field at every point on this surface?
    Ans : The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
  • Qstn #4
    A spherical conductor of radius 12 cm has a charge of 1.6 × 10-7C distributed uniformly on its surface. What is the electric field
  • #4-a
    Inside the sphere
    Ans : Radius of the spherical conductor, r = 12 cm = 0.12 m

    Charge is uniformly distributed over the conductor, q = 1.6 × 10-7 C

    Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
  • #4-b
    Just outside the sphere
    Ans : Electric field E just outside the conductor is given by the relation,



    Where,

    = Permittivity of free space



    Therefore, the electric field just outside the sphere is .
  • #4-c
    At a point 18 cm from the centre of the sphere?
    Ans : Electric field at a point 18 m from the centre of the sphere = E1

    Distance of the point from the centre, d = 18 cm = 0.18 m



    Therefore, the electric field at a point 18 cm from the centre of the sphere is

    .
  • Qstn #5
    A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
    Ans : Capacitance between the parallel plates of the capacitor, C = 8 pF

    Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1

    Capacitance, C, is given by the formula,



    Where,

    A = Area of each plate

    = Permittivity of free space

    If distance between the plates is reduced to half, then new distance, d’ =

    Dielectric constant of the substance filled in between the plates, = 6

    Hence, capacitance of the capacitor becomes



    Taking ratios of equations (i) and (ii), we obtain



    Therefore, the capacitance between the plates is 96 pF.
  • Qstn #6
    Three capacitors each of capacitance 9 pF are connected in series.
  • #6-a
    What is the total capacitance of the combination?
    Ans : Capacitance of each of the three capacitors, C = 9 pF

    Equivalent capacitance (C’) of the combination of the capacitors is given by the relation,

    1C’=1C+1C+1C⇒1C’=19+19+19=13⇒C’=3pFTherefore, total capacitance of the combination is

    3pF.
  • #6-b
    What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
    Ans : Supply voltage, V = 120 V

    Potential difference (V‘) across each capacitor is equal to one-third of the supply voltage.



    Therefore, the potential difference across each capacitor is 40 V.
  • Qstn #7
    Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
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