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Electrostatic Potential and Capacitance -CBSE Class 12th NCERT Solution

Jul 25, 2013 11:55 IST

    Get, detailed solutions to the questions of the chapter Electrostatic Potential and Capacitance from NCERT textbooks. The objective is to helping students regarding the pattern of answering the question as per the cbse latest marking scheme.Cbse.jagranjosh.com provided you NCERT solutions for classes 12th math and science subjects .

      Some questions of this chapter are given here

    Q. 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.


    Q1 = 5x10-8 C
    Q2 = 3x10-8 C
    r = 16 cm

    We know


    Let the point P be at distance x from Q1 and 16-x from Q2 where the electric potential is zero.
    Solving for cancellation of potential due to given charges

    Or,   (5 / x) + (3 / 16-x) = 0
    Or,    x = 40 cm from positive charge towards negative charge on extended line.
    Again, in between charges
    (5 / x) + (3 / x-16) = 0
    Or,   x = 10 cm from positive charge towards negative charge.

     Q.   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.


    Six charges = 5 µC
    Side of hexagon = 10 cm
    Distance between center to vertex = 10 cm
    Now, we know

    Substituting the given values
    V = (9x109).(5x10-6) / (10x10-2)
    Or,   V = 4.5x105 V

    Since all six charges are of equal magnitude and sign, therefore
    Net potential at the centre = 6x(4.5x105) = 2.7x106 V

    Q.   Two charges 2 μC and –2 μC are placed at points A and B 6 cm apart

    a)     Identify an equipotential surface of the system.

    b)    What is the direction of the electric field at every point on this surface?


    Q1 = 2 µC
    Q2 = -2 µC
    r = 6 cm

    a) Since both charges are equal and opposite, they will cancel out each other’s effect at the centre of line joining them, and the plane passing through it will have equal potential (i.e. zero).

    Normal to the plane in the direction AB.

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